From ccfa2c495fc5c7762517dc990a2fe58328a3292a Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E5=AD=9F=E4=BB=A4=E4=BB=A4?= Date: Fri, 8 Apr 2022 18:02:30 +0800 Subject: [PATCH 001/105] =?UTF-8?q?Update=200028.=E5=AE=9E=E7=8E=B0strStr.?= =?UTF-8?q?md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 添加 swift 前缀表两种实现方法 --- problems/0028.实现strStr.md | 107 ++++++++++++++++++++++++++++++++++ 1 file changed, 107 insertions(+) diff --git a/problems/0028.实现strStr.md b/problems/0028.实现strStr.md index 634d8535..d67e5f70 100644 --- a/problems/0028.实现strStr.md +++ b/problems/0028.实现strStr.md @@ -1059,5 +1059,112 @@ func getNext(_ next: inout [Int], needle: [Character]) { ``` +> 前缀表右移 + +```swift +func strStr(_ haystack: String, _ needle: String) -> Int { + + let s = Array(haystack), p = Array(needle) + guard p.count != 0 else { return 0 } + + var j = 0 + var next = [Int].init(repeating: 0, count: p.count) + getNext(&next, p) + + for i in 0 ..< s.count { + + while j > 0 && s[i] != p[j] { + j = next[j] + } + + if s[i] == p[j] { + j += 1 + } + + if j == p.count { + return i - p.count + 1 + } + } + + return -1 + } + + // 前缀表后移一位，首位用 -1 填充 + func getNext(_ next: inout [Int], _ needle: [Character]) { + + guard needle.count > 1 else { return } + + var j = 0 + next[0] = j + + for i in 1 ..< needle.count-1 { + + while j > 0 && needle[i] != needle[j] { + j = next[j-1] + } + + if needle[i] == needle[j] { + j += 1 + } + + next[i] = j + } + next.removeLast() + next.insert(-1, at: 0) + } +``` + +> 前缀表统一不减一 +```swift + +func strStr(_ haystack: String, _ needle: String) -> Int { + + let s = Array(haystack), p = Array(needle) + guard p.count != 0 else { return 0 } + + var j = 0 + var next = [Int](repeating: 0, count: needle.count) + // KMP + getNext(&next, needle: p) + + for i in 0 ..< s.count { + while j > 0 && s[i] != p[j] { + j = next[j-1] + } + + if s[i] == p[j] { + j += 1 + } + + if j == p.count { + return i - p.count + 1 + } + } + return -1 + } + + //前缀表 + func getNext(_ next: inout [Int], needle: [Character]) { + + var j = 0 + next[0] = j + + for i in 1 ..< needle.count { + + while j>0 && needle[i] != needle[j] { + j = next[j-1] + } + + if needle[i] == needle[j] { + j += 1 + } + + next[i] = j + + } + } + +``` + -----------------------
From 559f03c92869ca55ae929d66de872f5ddda70567 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E5=AD=9F=E4=BB=A4=E4=BB=A4?= Date: Fri, 8 Apr 2022 18:11:05 +0800 Subject: [PATCH 002/105] =?UTF-8?q?Update=200459.=E9=87=8D=E5=A4=8D?= =?UTF-8?q?=E7=9A=84=E5=AD=90=E5=AD=97=E7=AC=A6=E4=B8=B2.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0459.重复的子字符串.md | 43 ++++++++++++++++++++++++++ 1 file changed, 43 insertions(+) diff --git a/problems/0459.重复的子字符串.md b/problems/0459.重复的子字符串.md index ccfb485c..6a9b4260 100644 --- a/problems/0459.重复的子字符串.md +++ b/problems/0459.重复的子字符串.md @@ -421,5 +421,48 @@ function repeatedSubstringPattern(s: string): boolean { }; ``` + +Swift: + +> 前缀表统一减一 +```swift + func repeatedSubstringPattern(_ s: String) -> Bool { + + let sArr = Array(s) + let len = s.count + if len == 0 { + return false + } + var next = Array.init(repeating: -1, count: len) + + getNext(&next,sArr) + + if next.last != -1 && len % (len - (next[len-1] + 1)) == 0{ + return true + } + + return false + } + + func getNext(_ next: inout [Int], _ str:[Character]) { + + var j = -1 + next[0] = j + + for i in 1 ..< str.count { + + while j >= 0 && str[j+1] != str[i] { + j = next[j] + } + + if str[i] == str[j+1] { + j += 1 + } + + next[i] = j + } + } +``` + -----------------------
From 426b00a4e2889bb4c6d2599f29c965e173a5f536 Mon Sep 17 00:00:00 2001 From: Nada Date: Sun, 10 Apr 2022 14:56:56 +0800 Subject: [PATCH 003/105] =?UTF-8?q?Update=200018.=E5=9B=9B=E6=95=B0?= =?UTF-8?q?=E4=B9=8B=E5=92=8C.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 增加剪枝逻辑处理 --- problems/0018.四数之和.md | 15 ++++++++++----- 1 file changed, 10 insertions(+), 5 deletions(-) diff --git a/problems/0018.四数之和.md b/problems/0018.四数之和.md index 7304254e..ee70cb69 100644 --- a/problems/0018.四数之和.md +++ b/problems/0018.四数之和.md @@ -31,7 +31,7 @@ 四数之和，和[15.三数之和](https://programmercarl.com/0015.三数之和.html)是一个思路，都是使用双指针法, 基本解法就是在[15.三数之和](https://programmercarl.com/0015.三数之和.html) 的基础上再套一层for循环。 -但是有一些细节需要注意，例如： 不要判断`nums[k] > target` 就返回了，三数之和 可以通过 `nums[i] > 0` 就返回了，因为 0 已经是确定的数了，四数之和这道题目 target是任意值。（大家亲自写代码就能感受出来） +但是有一些细节需要注意，例如： 不要判断`nums[k] > target` 就返回了，三数之和 可以通过 `nums[i] > 0` 就返回了，因为 0 已经是确定的数了，四数之和这道题目 target是任意值。比如：数组是`[-4, -3, -2, -1]`，`target`是`-10`，不能因为`-4 > -10`而跳过。但是我们依旧可以去做剪枝，逻辑变成`nums[i] > target && (nums[i] >=0 || target >= 0)`就可以了。 [15.三数之和](https://programmercarl.com/0015.三数之和.html)的双指针解法是一层for循环num[i]为确定值，然后循环内有left和right下标作为双指针，找到nums[i] + nums[left] + nums[right] == 0。 @@ -72,15 +72,20 @@ public: vector> result; sort(nums.begin(), nums.end()); for (int k = 0; k < nums.size(); k++) { - // 这种剪枝是错误的，这道题目target 是任意值 - // if (nums[k] > target) { - // return result; - // } + // 剪枝处理 + if (nums[k] > target && (nums[k] >= 0 || target >= 0)) { + break; // 这里使用break，统一通过最后的return返回 + } // 去重 if (k > 0 && nums[k] == nums[k - 1]) { continue; } for (int i = k + 1; i < nums.size(); i++) { + // 2级剪枝处理 + if (nums[k] + nums[i] > target && (nums[k] + nums[i] >= 0 || target >= 0)) { + break; + } + // 正确去重方法 if (i > k + 1 && nums[i] == nums[i - 1]) { continue; From 0fc1cb334d4f8612e95eabfc9e97fe8d273dc25e Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E5=AD=9F=E4=BB=A4=E4=BB=A4?= Date: Wed, 13 Apr 2022 15:51:42 +0800 Subject: [PATCH 004/105] =?UTF-8?q?Update=200459.=E9=87=8D=E5=A4=8D?= =?UTF-8?q?=E7=9A=84=E5=AD=90=E5=AD=97=E7=AC=A6=E4=B8=B2.md=20=E6=B7=BB?= =?UTF-8?q?=E5=8A=A0swift=E6=96=B9=E6=B3=95(=E5=89=8D=E7=BC=80=E8=A1=A8?= =?UTF-8?q?=E7=BB=9F=E4=B8=80=E4=B8=8D=E5=87=8F=E4=B8=80)?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0459.重复的子字符串.md | 40 ++++++++++++++++++++++++++ 1 file changed, 40 insertions(+) diff --git a/problems/0459.重复的子字符串.md b/problems/0459.重复的子字符串.md index 6a9b4260..f2fe728d 100644 --- a/problems/0459.重复的子字符串.md +++ b/problems/0459.重复的子字符串.md @@ -463,6 +463,46 @@ Swift: } } ``` +> 前缀表统一不减一 +```swift + func repeatedSubstringPattern(_ s: String) -> Bool { + + let sArr = Array(s) + let len = sArr.count + if len == 0 { + return false + } + + var next = Array.init(repeating: 0, count: len) + getNext(&next, sArr) + + if next[len-1] != 0 && len % (len - next[len-1]) == 0 { + return true + } + + return false + } + + // 前缀表不减一 + func getNext(_ next: inout [Int], _ sArr:[Character]) { + + var j = 0 + next[0] = 0 + + for i in 1 ..< sArr.count { + while j > 0 && sArr[i] != sArr[j] { + j = next[j-1] + } + + if sArr[i] == sArr[j] { + j += 1 + } + + next[i] = j + } + } + +``` -----------------------
From cfb8ab8b936418f6b446bad655be491f9159be14 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E5=AD=9F=E4=BB=A4=E4=BB=A4?= Date: Wed, 13 Apr 2022 16:01:26 +0800 Subject: [PATCH 005/105] Revert "Merge branch 'youngyangyang04:master' into master" This reverts commit 33be177e63bc622c4e1e67d9e268bc13d4d5ba54, reversing changes made to 559f03c92869ca55ae929d66de872f5ddda70567. --- problems/0332.重新安排行程.md | 58 ----------------------------- 1 file changed, 58 deletions(-) diff --git a/problems/0332.重新安排行程.md b/problems/0332.重新安排行程.md index 041a7f03..01f81c4d 100644 --- a/problems/0332.重新安排行程.md +++ b/problems/0332.重新安排行程.md @@ -342,64 +342,6 @@ class Solution: return path ``` -### Go -```go -type pair struct { - target string - visited bool -} -type pairs []*pair - -func (p pairs) Len() int { - return len(p) -} -func (p pairs) Swap(i, j int) { - p[i], p[j] = p[j], p[i] -} -func (p pairs) Less(i, j int) bool { - return p[i].target < p[j].target -} - -func findItinerary(tickets [][]string) []string { - result := []string{} - // map[出发机场] pair{目的地,是否被访问过} - targets := make(map[string]pairs) - for _, ticket := range tickets { - if targets[ticket[0]] == nil { - targets[ticket[0]] = make(pairs, 0) - } - targets[ticket[0]] = append(targets[ticket[0]], &pair{target: ticket[1], visited: false}) - } - for k, _ := range targets { - sort.Sort(targets[k]) - } - result = append(result, "JFK") - var backtracking func() bool - backtracking = func() bool { - if len(tickets)+1 == len(result) { - return true - } - // 取出起飞航班对应的目的地 - for _, pair := range targets[result[len(result)-1]] { - if pair.visited == false { - result = append(result, pair.target) - pair.visited = true - if backtracking() { - return true - } - result = result[:len(result)-1] - pair.visited = false - } - } - return false - } - - backtracking() - - return result -} -``` - ### C语言 ```C From efe987be6b6a9cdc908d836e86954aab22227da7 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E5=AD=9F=E4=BB=A4=E4=BB=A4?= Date: Wed, 13 Apr 2022 16:07:23 +0800 Subject: [PATCH 006/105] =?UTF-8?q?Revert=20"Update=200459.=E9=87=8D?= =?UTF-8?q?=E5=A4=8D=E7=9A=84=E5=AD=90=E5=AD=97=E7=AC=A6=E4=B8=B2.md=20?= =?UTF-8?q?=E6=B7=BB=E5=8A=A0swift=E6=96=B9=E6=B3=95(=E5=89=8D=E7=BC=80?= =?UTF-8?q?=E8=A1=A8=E7=BB=9F=E4=B8=80=E4=B8=8D=E5=87=8F=E4=B8=80)"?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit This reverts commit 0fc1cb334d4f8612e95eabfc9e97fe8d273dc25e. --- problems/0459.重复的子字符串.md | 40 -------------------------- 1 file changed, 40 deletions(-) diff --git a/problems/0459.重复的子字符串.md b/problems/0459.重复的子字符串.md index f2fe728d..6a9b4260 100644 --- a/problems/0459.重复的子字符串.md +++ b/problems/0459.重复的子字符串.md @@ -463,46 +463,6 @@ Swift: } } ``` -> 前缀表统一不减一 -```swift - func repeatedSubstringPattern(_ s: String) -> Bool { - - let sArr = Array(s) - let len = sArr.count - if len == 0 { - return false - } - - var next = Array.init(repeating: 0, count: len) - getNext(&next, sArr) - - if next[len-1] != 0 && len % (len - next[len-1]) == 0 { - return true - } - - return false - } - - // 前缀表不减一 - func getNext(_ next: inout [Int], _ sArr:[Character]) { - - var j = 0 - next[0] = 0 - - for i in 1 ..< sArr.count { - while j > 0 && sArr[i] != sArr[j] { - j = next[j-1] - } - - if sArr[i] == sArr[j] { - j += 1 - } - - next[i] = j - } - } - -``` -----------------------
From a391c2c949c393141e7aed3d1e8cdbf6b6a0b956 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E5=AD=9F=E4=BB=A4=E4=BB=A4?= Date: Wed, 13 Apr 2022 16:10:50 +0800 Subject: [PATCH 007/105] =?UTF-8?q?Update=200459.=E9=87=8D=E5=A4=8D?= =?UTF-8?q?=E7=9A=84=E5=AD=90=E5=AD=97=E7=AC=A6=E4=B8=B2.md=20=E6=B7=BB?= =?UTF-8?q?=E5=8A=A0swift=E6=96=B9=E6=B3=95(=E5=89=8D=E7=BC=80=E8=A1=A8?= =?UTF-8?q?=E7=BB=9F=E4=B8=80=E4=B8=8D=E5=87=8F=E4=B8=80)?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0459.重复的子字符串.md | 41 ++++++++++++++++++++++++++ 1 file changed, 41 insertions(+) diff --git a/problems/0459.重复的子字符串.md b/problems/0459.重复的子字符串.md index 6a9b4260..a51c68ee 100644 --- a/problems/0459.重复的子字符串.md +++ b/problems/0459.重复的子字符串.md @@ -464,5 +464,46 @@ Swift: } ``` +> 前缀表统一不减一 +```swift + func repeatedSubstringPattern(_ s: String) -> Bool { + + let sArr = Array(s) + let len = sArr.count + if len == 0 { + return false + } + + var next = Array.init(repeating: 0, count: len) + getNext(&next, sArr) + + if next[len-1] != 0 && len % (len - next[len-1]) == 0 { + return true + } + + return false + } + + // 前缀表不减一 + func getNext(_ next: inout [Int], _ sArr:[Character]) { + + var j = 0 + next[0] = 0 + + for i in 1 ..< sArr.count { + + while j > 0 && sArr[i] != sArr[j] { + j = next[j-1] + } + + if sArr[i] == sArr[j] { + j += 1 + } + + next[i] = j + } + } +``` + -----------------------
From 7c26aba0e1d9ba38a631eb5dbae8577890deb2d2 Mon Sep 17 00:00:00 2001 From: Falldio <1031249495@qq.com> Date: Sun, 17 Apr 2022 11:59:30 +0800 Subject: [PATCH 008/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880209.?= =?UTF-8?q?=E9=95=BF=E5=BA=A6=E6=9C=80=E5=B0=8F=E7=9A=84=E5=AD=90=E6=95=B0?= =?UTF-8?q?=E7=BB=84.md=EF=BC=89=EF=BC=9A=E5=A2=9E=E5=8A=A0Kotlin=E7=89=88?= =?UTF-8?q?=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0209.长度最小的子数组.md | 31 +++++++++++++++++++---- 1 file changed, 26 insertions(+), 5 deletions(-) diff --git a/problems/0209.长度最小的子数组.md b/problems/0209.长度最小的子数组.md index 82a11381..fd72cf1b 100644 --- a/problems/0209.长度最小的子数组.md +++ b/problems/0209.长度最小的子数组.md @@ -198,7 +198,7 @@ JavaScript: var minSubArrayLen = function(target, nums) { // 长度计算一次 const len = nums.length; - let l = r = sum = 0, + let l = r = sum = 0, res = len + 1; // 子数组最大不会超过自身 while(r < len) { sum += nums[r++]; @@ -260,12 +260,12 @@ Rust: ```rust impl Solution { - pub fn min_sub_array_len(target: i32, nums: Vec) -> i32 { + pub fn min_sub_array_len(target: i32, nums: Vec) -> i32 { let (mut result, mut subLength): (i32, i32) = (i32::MAX, 0); let (mut sum, mut i) = (0, 0); - + for (pos, val) in nums.iter().enumerate() { - sum += val; + sum += val; while sum >= target { subLength = (pos - i + 1) as i32; if result > subLength { @@ -364,7 +364,7 @@ int minSubArrayLen(int target, int* nums, int numsSize){ int minLength = INT_MAX; int sum = 0; - int left = 0, right = 0; + int left = 0, right = 0; //右边界向右扩展 for(; right < numsSize; ++right) { sum += nums[right]; @@ -380,5 +380,26 @@ int minSubArrayLen(int target, int* nums, int numsSize){ } ``` +Kotlin: +```kotlin +class Solution { + fun minSubArrayLen(target: Int, nums: IntArray): Int { + var start = 0 + var end = 0 + var ret = Int.MAX_VALUE + var count = 0 + while (end < nums.size) { + count += nums[end] + while (count >= target) { + ret = if (ret > (end - start + 1)) end - start + 1 else ret + count -= nums[start++] + } + end++ + } + return if (ret == Int.MAX_VALUE) 0 else ret + } +} +``` + -----------------------
From 0375a5b147c40a449c04ef7f988dfcb2d93b9e01 Mon Sep 17 00:00:00 2001 From: Guacker <92098421+Guacker@users.noreply.github.com> Date: Sun, 17 Apr 2022 17:19:00 +0800 Subject: [PATCH 009/105] =?UTF-8?q?Update=20=E4=BA=8C=E5=8F=89=E6=A0=91?= =?UTF-8?q?=E7=9A=84=E9=80=92=E5=BD=92=E9=81=8D=E5=8E=86.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 修改了c语言实现的先序遍历函数名 和内部调用时用的函数名 及在preorderTraversal中的调用先序遍历操作的函数名 --- problems/二叉树的递归遍历.md | 8 ++++---- 1 file changed, 4 insertions(+), 4 deletions(-) diff --git a/problems/二叉树的递归遍历.md b/problems/二叉树的递归遍历.md index 35d19d7b..612f2394 100644 --- a/problems/二叉树的递归遍历.md +++ b/problems/二叉树的递归遍历.md @@ -371,18 +371,18 @@ C: ```c //前序遍历： -void preOrderTraversal(struct TreeNode* root, int* ret, int* returnSize) { +void preOrder(struct TreeNode* root, int* ret, int* returnSize) { if(root == NULL) return; ret[(*returnSize)++] = root->val; - preOrderTraverse(root->left, ret, returnSize); - preOrderTraverse(root->right, ret, returnSize); + preOrder(root->left, ret, returnSize); + preOrder(root->right, ret, returnSize); } int* preorderTraversal(struct TreeNode* root, int* returnSize){ int* ret = (int*)malloc(sizeof(int) * 100); *returnSize = 0; - preOrderTraversal(root, ret, returnSize); + preOrder(root, ret, returnSize); return ret; } From db84840986aab5ce791230eba963084bf1d5ba4b Mon Sep 17 00:00:00 2001 From: GitHubQAQ <31883473+GitHubQAQ@users.noreply.github.com> Date: Tue, 19 Apr 2022 14:23:08 +0800 Subject: [PATCH 010/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=E9=A2=84=E5=88=A4?= =?UTF-8?q?=E6=96=AD=E9=95=BF=E5=BA=A6=E7=9A=84=E4=BB=A3=E7=A0=81?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 在magazine中字符存入数组前，预先判断ransomNote和magazine的相对长度，进行处理。 --- problems/0383.赎金信.md | 4 ++++ 1 file changed, 4 insertions(+) diff --git a/problems/0383.赎金信.md b/problems/0383.赎金信.md index 00707347..5d9e8295 100644 --- a/problems/0383.赎金信.md +++ b/problems/0383.赎金信.md @@ -84,6 +84,10 @@ class Solution { public: bool canConstruct(string ransomNote, string magazine) { int record[26] = {0}; + //add + if (ransomNote.size() > magazine.size()) { + return false; + } for (int i = 0; i < magazine.length(); i++) { // 通过recode数据记录 magazine里各个字符出现次数 record[magazine[i]-'a'] ++; From 68e47dc4b30ffe38d1a26bb5104614f2f1cc067a Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Tue, 19 Apr 2022 17:10:20 +0800 Subject: [PATCH 011/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880452.?= =?UTF-8?q?=E7=94=A8=E6=9C=80=E5=B0=91=E6=95=B0=E9=87=8F=E7=9A=84=E7=AE=AD?= =?UTF-8?q?=E5=BC=95=E7=88=86=E6=B0=94=E7=90=83.md=EF=BC=89=EF=BC=9A?= =?UTF-8?q?=E5=A2=9E=E5=8A=A0typescript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../0452.用最少数量的箭引爆气球.md | 26 ++++++++++++++++++- 1 file changed, 25 insertions(+), 1 deletion(-) diff --git a/problems/0452.用最少数量的箭引爆气球.md b/problems/0452.用最少数量的箭引爆气球.md index 33bbad55..2ab14b61 100644 --- a/problems/0452.用最少数量的箭引爆气球.md +++ b/problems/0452.用最少数量的箭引爆气球.md @@ -193,7 +193,7 @@ func min(a,b int) int{ } return a } -``` +``` ### Javascript ```Javascript @@ -214,7 +214,31 @@ var findMinArrowShots = function(points) { }; ``` +### TypeScript + +```typescript +function findMinArrowShots(points: number[][]): number { + const length: number = points.length; + if (length === 0) return 0; + points.sort((a, b) => a[0] - b[0]); + let resCount: number = 1; + let right: number = points[0][1]; // 右边界 + let tempPoint: number[]; + for (let i = 1; i < length; i++) { + tempPoint = points[i]; + if (tempPoint[0] > right) { + resCount++; + right = tempPoint[1]; + } else { + right = Math.min(right, tempPoint[1]); + } + } + return resCount; +}; +``` + ### C + ```c int cmp(const void *a,const void *b) { From f9146a3b98fd46ca548c8852bf3b7df0625cfe72 Mon Sep 17 00:00:00 2001 From: zhenghao <1650937065@qq.com> Date: Wed, 20 Apr 2022 10:23:18 +0800 Subject: [PATCH 012/105] =?UTF-8?q?=E6=9B=B4=E6=96=B00435=20=E6=97=A0?= =?UTF-8?q?=E9=87=8D=E5=8F=A0=E5=8C=BA=E9=97=B4=20java=E7=89=88=E6=9C=AC?= =?UTF-8?q?=E5=8F=B3=E8=BE=B9=E7=95=8C=E8=A7=A3=E6=B3=95=E4=B8=AD=E6=8E=92?= =?UTF-8?q?=E5=BA=8F=E7=9A=84=E5=88=A4=E6=96=AD=E9=80=BB=E8=BE=91?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0435.无重叠区间.md | 5 +++-- 1 file changed, 3 insertions(+), 2 deletions(-) diff --git a/problems/0435.无重叠区间.md b/problems/0435.无重叠区间.md index b24ca024..cf58b9e2 100644 --- a/problems/0435.无重叠区间.md +++ b/problems/0435.无重叠区间.md @@ -184,13 +184,14 @@ public: class Solution { public int eraseOverlapIntervals(int[][] intervals) { Arrays.sort(intervals, (a, b) -> { - if (a[0] == a[0]) return a[1] - b[1]; - return a[0] - b[0]; + // 按照区间右边界升序排序 + return a[1] - b[1]; }); int count = 0; int edge = Integer.MIN_VALUE; for (int i = 0; i < intervals.length; i++) { + // 若上一个区间的右边界小于当前区间的左边界，说明无交集 if (edge <= intervals[i][0]) { edge = intervals[i][1]; } else { From b5f11af3edc66af674d486ace76cde8a99637170 Mon Sep 17 00:00:00 2001 From: zhenyu <48195906+szluyu99@users.noreply.github.com> Date: Wed, 20 Apr 2022 23:56:20 +0800 Subject: [PATCH 013/105] =?UTF-8?q?Update=20=E8=83=8C=E5=8C=85=E9=97=AE?= =?UTF-8?q?=E9=A2=98=E7=90=86=E8=AE=BA=E5=9F=BA=E7=A1=80=E5=AE=8C=E5=85=A8?= =?UTF-8?q?=E8=83=8C=E5=8C=85.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/背包问题理论基础完全背包.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/背包问题理论基础完全背包.md b/problems/背包问题理论基础完全背包.md index faa1dc46..3ec399f1 100644 --- a/problems/背包问题理论基础完全背包.md +++ b/problems/背包问题理论基础完全背包.md @@ -37,7 +37,7 @@ * [动态规划：关于01背包问题，你该了解这些！（滚动数组）](https://programmercarl.com/背包理论基础01背包-2.html) 首先在回顾一下01背包的核心代码 -``` +```cpp for(int i = 0; i < weight.size(); i++) { // 遍历物品 for(int j = bagWeight; j >= weight[i]; j--) { // 遍历背包容量 dp[j] = max(dp[j], dp[j - weight[i]] + value[i]); From 4154d3db2eadecdddaba45d09f09bfe16c0cc594 Mon Sep 17 00:00:00 2001 From: 243wresfdxvc Date: Thu, 21 Apr 2022 09:55:19 +0000 Subject: [PATCH 014/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200134.=20=E5=8A=A0?= =?UTF-8?q?=E6=B2=B9=E7=AB=99.md=20C=E8=AF=AD=E8=A8=80=E8=B4=AA=E5=BF=83?= =?UTF-8?q?=E8=A7=A3=E6=B3=95=E6=96=B9=E6=B3=95=E4=BA=8C?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0134.加油站.md | 32 ++++++++++++++++++++++++++++++++ 1 file changed, 32 insertions(+) diff --git a/problems/0134.加油站.md b/problems/0134.加油站.md index ca95af67..45e05fed 100644 --- a/problems/0134.加油站.md +++ b/problems/0134.加油站.md @@ -341,6 +341,7 @@ var canCompleteCircuit = function(gas, cost) { ``` ### C +贪心算法：方法一 ```c int canCompleteCircuit(int* gas, int gasSize, int* cost, int costSize){ int curSum = 0; @@ -370,5 +371,36 @@ int canCompleteCircuit(int* gas, int gasSize, int* cost, int costSize){ } ``` +贪心算法：方法二 +```c +int canCompleteCircuit(int* gas, int gasSize, int* cost, int costSize){ + int curSum = 0; + int totalSum = 0; + int start = 0; + + int i; + for(i = 0; i < gasSize; ++i) { + // 当前i站中加油量与耗油量的差 + int diff = gas[i] - cost[i]; + + curSum += diff; + totalSum += diff; + + // 若0到i的加油量都为负，则开始位置应为i+1 + if(curSum < 0) { + curSum = 0; + // 当i + 1 == gasSize时，totalSum < 0（此时i为gasSize - 1)，油车不可能返回原点 + start = i + 1; + } + } + + // 若总和小于0，加油车无论如何都无法返回原点。返回-1 + if(totalSum < 0) + return -1; + + return start; +} +``` + -----------------------
From be07599474b87bf7dba3e97107b19435a304e5e7 Mon Sep 17 00:00:00 2001 From: berserk-112 <40333359+berserk-112@users.noreply.github.com> Date: Fri, 22 Apr 2022 09:16:34 +0800 Subject: [PATCH 015/105] =?UTF-8?q?Update=200309.=E6=9C=80=E4=BD=B3?= =?UTF-8?q?=E4=B9=B0=E5=8D=96=E8=82=A1=E7=A5=A8=E6=97=B6=E6=9C=BA=E5=90=AB?= =?UTF-8?q?=E5=86=B7=E5=86=BB=E6=9C=9F.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0309.最佳买卖股票时机含冷冻期.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/problems/0309.最佳买卖股票时机含冷冻期.md b/problems/0309.最佳买卖股票时机含冷冻期.md index 53caa46e..8bd2fcf8 100644 --- a/problems/0309.最佳买卖股票时机含冷冻期.md +++ b/problems/0309.最佳买卖股票时机含冷冻期.md @@ -214,8 +214,8 @@ class Solution { for (int i = 2; i <= prices.length; i++) { /* - dp[i][0] 第i天未持有股票收益; - dp[i][1] 第i天持有股票收益; + dp[i][0] 第i天持有股票收益; + dp[i][1] 第i天不持有股票收益; 情况一：第i天是冷静期，不能以dp[i-1][1]购买股票,所以以dp[i - 2][1]买股票，没问题 情况二：第i天不是冷静期，理论上应该以dp[i-1][1]购买股票，但是第i天不是冷静期说明，第i-1天没有卖出股票， 则dp[i-1][1]=dp[i-2][1],所以可以用dp[i-2][1]买股票，没问题 From c3f746a25d81c299ca8a3efa4fc1f9e8d45509d3 Mon Sep 17 00:00:00 2001 From: qyg <1600314850@qq.com> Date: Fri, 22 Apr 2022 10:55:01 +0800 Subject: [PATCH 016/105] =?UTF-8?q?0209.=E9=95=BF=E5=BA=A6=E6=9C=80?= =?UTF-8?q?=E5=B0=8F=E7=9A=84=E5=AD=90=E6=95=B0=E7=BB=84=EF=BC=9A=E8=B0=83?= =?UTF-8?q?=E6=95=B4=E7=AC=94=E8=AF=AF?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0209.长度最小的子数组.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0209.长度最小的子数组.md b/problems/0209.长度最小的子数组.md index 82a11381..23b27edd 100644 --- a/problems/0209.长度最小的子数组.md +++ b/problems/0209.长度最小的子数组.md @@ -112,7 +112,7 @@ public: **一些录友会疑惑为什么时间复杂度是O(n)**。 -不要以为for里放一个while就以为是O(n^2)啊， 主要是看每一个元素被操作的次数，每个元素在滑动窗后进来操作一次，出去操作一次，每个元素都是被被操作两次，所以时间复杂度是 2 × n 也就是O(n)。 +不要以为for里放一个while就以为是O(n^2)啊， 主要是看每一个元素被操作的次数，每个元素在滑动窗后进来操作一次，出去操作一次，每个元素都是被操作两次，所以时间复杂度是 2 × n 也就是O(n)。 ## 相关题目推荐 From 4d668b0efc89cd15f50dcb93da01e1ace633b07b Mon Sep 17 00:00:00 2001 From: qyg <1600314850@qq.com> Date: Fri, 22 Apr 2022 11:02:11 +0800 Subject: [PATCH 017/105] =?UTF-8?q?0059.=E8=9E=BA=E6=97=8B=E7=9F=A9?= =?UTF-8?q?=E9=98=B5II=EF=BC=9A=E8=B0=83=E6=95=B4=E7=AC=94=E8=AF=AF?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0059.螺旋矩阵II.md | 6 +++--- 1 file changed, 3 insertions(+), 3 deletions(-) diff --git a/problems/0059.螺旋矩阵II.md b/problems/0059.螺旋矩阵II.md index 5c679982..1162c7eb 100644 --- a/problems/0059.螺旋矩阵II.md +++ b/problems/0059.螺旋矩阵II.md @@ -30,7 +30,7 @@ 相信很多同学刚开始做这种题目的时候，上来就是一波判断猛如虎。 -结果运行的时候各种问题，然后开始各种修修补补，最后发现改了这里哪里有问题，改了那里这里又跑不起来了。 +结果运行的时候各种问题，然后开始各种修修补补，最后发现改了这里那里有问题，改了那里这里又跑不起来了。 大家还记得我们在这篇文章[数组：每次遇到二分法，都是一看就会，一写就废](https://programmercarl.com/0704.二分查找.html)中讲解了二分法，提到如果要写出正确的二分法一定要坚持**循环不变量原则**。 @@ -47,7 +47,7 @@ 可以发现这里的边界条件非常多，在一个循环中，如此多的边界条件，如果不按照固定规则来遍历，那就是**一进循环深似海，从此offer是路人**。 -这里一圈下来，我们要画每四条边，这四条边怎么画，每画一条边都要坚持一致的左闭右开，或者左开又闭的原则，这样这一圈才能按照统一的规则画下来。 +这里一圈下来，我们要画每四条边，这四条边怎么画，每画一条边都要坚持一致的左闭右开，或者左开右闭的原则，这样这一圈才能按照统一的规则画下来。 那么我按照左闭右开的原则，来画一圈，大家看一下： @@ -59,7 +59,7 @@ 一些同学做这道题目之所以一直写不好，代码越写越乱。 -就是因为在画每一条边的时候，一会左开又闭，一会左闭右闭，一会又来左闭右开，岂能不乱。 +就是因为在画每一条边的时候，一会左开右闭，一会左闭右闭，一会又来左闭右开，岂能不乱。 代码如下，已经详细注释了每一步的目的，可以看出while循环里判断的情况是很多的，代码里处理的原则也是统一的左闭右开。 From 664746fe155a137f01c5d1478b0325c2dd8e0ffe Mon Sep 17 00:00:00 2001 From: qyg <1600314850@qq.com> Date: Fri, 22 Apr 2022 11:21:18 +0800 Subject: [PATCH 018/105] =?UTF-8?q?=E6=95=B0=E7=BB=84=E6=80=BB=E7=BB=93?= =?UTF-8?q?=E7=AF=87=EF=BC=9A=E8=B0=83=E6=95=B4=E7=AC=94=E8=AF=AF=EF=BC=8C?= =?UTF-8?q?=E5=A2=9E=E5=8A=A0=E4=BA=8C=E7=BB=B4=E6=95=B0=E7=BB=84=E7=9A=84?= =?UTF-8?q?=E5=86=85=E5=AE=B9=E6=98=AF=E9=92=88=E5=AF=B9Java=E8=AF=AD?= =?UTF-8?q?=E8=A8=80=E7=9A=84=E9=99=90=E5=AE=9A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/数组总结篇.md | 8 ++++---- 1 file changed, 4 insertions(+), 4 deletions(-) diff --git a/problems/数组总结篇.md b/problems/数组总结篇.md index d256298b..39fa17a6 100644 --- a/problems/数组总结篇.md +++ b/problems/数组总结篇.md @@ -43,19 +43,19 @@ **那么二维数组在内存的空间地址是连续的么？** -我们来举一个例子，例如： `int[][] rating = new int[3][4];` ， 这个二维数据在内存空间可不是一个 `3*4` 的连续地址空间 +我们来举一个Java的例子，例如： `int[][] rating = new int[3][4];` ， 这个二维数组在内存空间可不是一个 `3*4` 的连续地址空间 看了下图，就应该明白了： -所以**二维数据在内存中不是 `3*4` 的连续地址空间，而是四条连续的地址空间组成！** +所以**Java的二维数组在内存中不是 `3*4` 的连续地址空间，而是四条连续的地址空间组成！** # 数组的经典题目 在面试中，数组是必考的基础数据结构。 -其实数据的题目在思想上一般比较简单的，但是如果想高效，并不容易。 +其实数组的题目在思想上一般比较简单的，但是如果想高效，并不容易。 我们之前一共讲解了四道经典数组题目，每一道题目都代表一个类型，一种思想。 @@ -115,7 +115,7 @@ 在这道题目中，我们再一次介绍到了**循环不变量原则**，其实这也是写程序中的重要原则。 -相信大家又遇到过这种情况： 感觉题目的边界调节超多，一波接着一波的判断，找边界，踩了东墙补西墙，好不容易运行通过了，代码写的十分冗余，毫无章法，其实**真正解决题目的代码都是简洁的，或者有原则性的**，大家可以在这道题目中体会到这一点。 +相信大家有遇到过这种情况： 感觉题目的边界调节超多，一波接着一波的判断，找边界，拆了东墙补西墙，好不容易运行通过了，代码写的十分冗余，毫无章法，其实**真正解决题目的代码都是简洁的，或者有原则性的**，大家可以在这道题目中体会到这一点。 # 总结 From 074937378aad536271afd42fdb5cc27a1f193f7f Mon Sep 17 00:00:00 2001 From: qyg <1600314850@qq.com> Date: Fri, 22 Apr 2022 11:42:50 +0800 Subject: [PATCH 019/105] =?UTF-8?q?=E9=93=BE=E8=A1=A8=E7=90=86=E8=AE=BA?= =?UTF-8?q?=E5=9F=BA=E7=A1=80=EF=BC=9A=E8=B0=83=E6=95=B4=E7=AC=94=E8=AF=AF?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/链表理论基础.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/problems/链表理论基础.md b/problems/链表理论基础.md index 095282f5..2fe9f14c 100644 --- a/problems/链表理论基础.md +++ b/problems/链表理论基础.md @@ -24,7 +24,7 @@ ## 双链表 -单链表中的节点只能指向节点的下一个节点。 +单链表中的指针域只能指向节点的下一个节点。 双链表：每一个节点有两个指针域，一个指向下一个节点，一个指向上一个节点。 @@ -56,7 +56,7 @@ ![链表3](https://img-blog.csdnimg.cn/20200806194613920.png) -这个链表起始节点为2， 终止节点为7， 各个节点分布在内存个不同地址空间上，通过指针串联在一起。 +这个链表起始节点为2， 终止节点为7， 各个节点分布在内存的不同地址空间上，通过指针串联在一起。 # 链表的定义 From 5a7e247e5f5a70c1e33dfce4cc42b498cc031f61 Mon Sep 17 00:00:00 2001 From: ChubbyPan Date: Fri, 22 Apr 2022 07:59:32 +0000 Subject: [PATCH 020/105] update build binary tree in ACM pattern with python --- .../前序/ACM模式如何构建二叉树.md | 56 ++++++++++++++++++- 1 file changed, 55 insertions(+), 1 deletion(-) diff --git a/problems/前序/ACM模式如何构建二叉树.md b/problems/前序/ACM模式如何构建二叉树.md index bd2e9780..f6ec2dd3 100644 --- a/problems/前序/ACM模式如何构建二叉树.md +++ b/problems/前序/ACM模式如何构建二叉树.md @@ -213,7 +213,61 @@ int main() { ## Python -```Python +```Python3 +class TreeNode: + def __init__(self, val = 0, left = None, right = None): + self.val = val + self.left = left + self.right = right + + +# 根据数组构建二叉树 + +def construct_binary_tree(nums: []) -> TreeNode: + if not nums: + return None + # 用于存放构建好的节点 + root = TreeNode(-1) + Tree = [] + # 将数组元素全部转化为树节点 + for i in range(len(nums)): + if nums[i]!= -1: + node = TreeNode(nums[i]) + else: + node = None + Tree.append(node) + if i == 0: + root = node + for i in range(len(Tree)): + node = Tree[i] + if node and (2 * i + 2) < len(Tree): + node.left = Tree[i * 2 + 1] + node.right = Tree[i * 2 + 2] + return root + + + +# 算法:中序遍历二叉树 + +class Solution: + def __init__(self): + self.T = [] + def inorder(self, root: TreeNode) -> []: + if not root: + return + self.inorder(root.left) + self.T.append(root.val) + self.inorder(root.right) + return self.T + + + +# 验证创建二叉树的有效性,二叉排序树的中序遍历应为顺序排列 + +test_tree = [3, 1, 5, -1, 2, 4 ,6] +root = construct_binary_tree(test_tree) +A = Solution() +print(A.inorder(root)) ``` From a0c16d48cb851b3c118cf3da0900797be767f9d0 Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Sat, 23 Apr 2022 18:19:36 +0800 Subject: [PATCH 021/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880435.?= =?UTF-8?q?=E6=97=A0=E9=87=8D=E5=8F=A0=E5=8C=BA=E9=97=B4.md=EF=BC=89?= =?UTF-8?q?=EF=BC=9A=E5=A2=9E=E5=8A=A0typescript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0435.无重叠区间.md | 51 +++++++++++++++++++++++++++++++- 1 file changed, 50 insertions(+), 1 deletion(-) diff --git a/problems/0435.无重叠区间.md b/problems/0435.无重叠区间.md index b24ca024..c7f755bf 100644 --- a/problems/0435.无重叠区间.md +++ b/problems/0435.无重叠区间.md @@ -263,7 +263,7 @@ func min(a,b int)int{ } return a } -``` +``` ### Javascript: - 按右边界排序 @@ -306,6 +306,55 @@ var eraseOverlapIntervals = function(intervals) { } ``` +### TypeScript + +> 按右边界排序，从左往右遍历 + +```typescript +function eraseOverlapIntervals(intervals: number[][]): number { + const length = intervals.length; + if (length === 0) return 0; + intervals.sort((a, b) => a[1] - b[1]); + let right: number = intervals[0][1]; + let count: number = 1; + for (let i = 1; i < length; i++) { + if (intervals[i][0] >= right) { + count++; + right = intervals[i][1]; + } + } + return length - count; +}; +``` + +> 按左边界排序，从左往右遍历 + +```typescript +function eraseOverlapIntervals(intervals: number[][]): number { + if (intervals.length === 0) return 0; + intervals.sort((a, b) => a[0] - b[0]); + let right: number = intervals[0][1]; + let tempInterval: number[]; + let resCount: number = 0; + for (let i = 1, length = intervals.length; i < length; i++) { + tempInterval = intervals[i]; + if (tempInterval[0] >= right) { + // 未重叠 + right = tempInterval[1]; + } else { + // 有重叠，移除当前interval和前一个interval中右边界更大的那个 + right = Math.min(right, tempInterval[1]); + resCount++; + } + } + return resCount; +}; +``` + + + + + -----------------------
From 7c752afaf472f8e3056d9bfefd9628548cbec5d8 Mon Sep 17 00:00:00 2001 From: h4 <20080114+tan-i-ham@users.noreply.github.com> Date: Sat, 23 Apr 2022 22:19:09 +0900 Subject: [PATCH 022/105] chore: Sync 150 markdown render format --- problems/0150.逆波兰表达式求值.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/problems/0150.逆波兰表达式求值.md b/problems/0150.逆波兰表达式求值.md index f4dad823..fd3d69aa 100644 --- a/problems/0150.逆波兰表达式求值.md +++ b/problems/0150.逆波兰表达式求值.md @@ -109,7 +109,7 @@ public: }; ``` -# 题外话 +## 题外话 我们习惯看到的表达式都是中缀表达式，因为符合我们的习惯，但是中缀表达式对于计算机来说就不是很友好了。 @@ -128,7 +128,7 @@ public: -# 其他语言版本 +## 其他语言版本 java: From 531c1b0a3bfac9867a8403baaf95a6cd284f6a41 Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Sat, 23 Apr 2022 21:19:23 +0800 Subject: [PATCH 023/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880763.?= =?UTF-8?q?=E5=88=92=E5=88=86=E5=AD=97=E6=AF=8D=E5=8C=BA=E9=97=B4.md?= =?UTF-8?q?=EF=BC=89=EF=BC=9A=E5=A2=9E=E5=8A=A0typescript=E7=89=88?= =?UTF-8?q?=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0763.划分字母区间.md | 25 +++++++++++++++++++++++++ 1 file changed, 25 insertions(+) diff --git a/problems/0763.划分字母区间.md b/problems/0763.划分字母区间.md index 03d3a73b..901dccb4 100644 --- a/problems/0763.划分字母区间.md +++ b/problems/0763.划分字母区间.md @@ -174,6 +174,31 @@ var partitionLabels = function(s) { }; ``` +### TypeScript + +```typescript +function partitionLabels(s: string): number[] { + const length: number = s.length; + const resArr: number[] = []; + const helperMap: Map = new Map(); + for (let i = 0; i < length; i++) { + helperMap.set(s[i], i); + } + let left: number = 0; + let right: number = 0; + for (let i = 0; i < length; i++) { + right = Math.max(helperMap.get(s[i])!, right); + if (i === right) { + resArr.push(i - left + 1); + left = i + 1; + } + } + return resArr; +}; +``` + + + -----------------------
From 8966752197c174e46052ab549495a045ce1a5f14 Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Sat, 23 Apr 2022 21:44:11 +0800 Subject: [PATCH 024/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880056.?= =?UTF-8?q?=E5=90=88=E5=B9=B6=E5=8C=BA=E9=97=B4.md=EF=BC=89=EF=BC=9A?= =?UTF-8?q?=E5=A2=9E=E5=8A=A0typescript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0056.合并区间.md | 20 ++++++++++++++++++++ 1 file changed, 20 insertions(+) diff --git a/problems/0056.合并区间.md b/problems/0056.合并区间.md index a9caeaf0..b44d602c 100644 --- a/problems/0056.合并区间.md +++ b/problems/0056.合并区间.md @@ -266,6 +266,26 @@ var merge = function(intervals) { }; ``` +### TypeScript + +```typescript +function merge(intervals: number[][]): number[][] { + const resArr: number[][] = []; + intervals.sort((a, b) => a[0] - b[0]); + resArr[0] = [...intervals[0]]; // 避免修改原intervals + for (let i = 1, length = intervals.length; i < length; i++) { + let interval: number[] = intervals[i]; + let last: number[] = resArr[resArr.length - 1]; + if (interval[0] <= last[1]) { + last[1] = Math.max(interval[1], last[1]); + } else { + resArr.push([...intervals[i]]); + } + } + return resArr; +}; +``` + ----------------------- From e2807eb59ad5b9e45311544278c905fd942e85b3 Mon Sep 17 00:00:00 2001 From: 243wresfdxvc Date: Sat, 23 Apr 2022 16:15:53 +0000 Subject: [PATCH 025/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200203.=E7=A7=BB?= =?UTF-8?q?=E5=87=BA=E9=93=BE=E8=A1=A8=E5=85=83=E7=B4=A0.md=20C=E8=AF=AD?= =?UTF-8?q?=E8=A8=80=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0203.移除链表元素.md | 32 +++++++++++++++++++++++++++++ 1 file changed, 32 insertions(+) diff --git a/problems/0203.移除链表元素.md b/problems/0203.移除链表元素.md index c34831b7..751553e2 100644 --- a/problems/0203.移除链表元素.md +++ b/problems/0203.移除链表元素.md @@ -145,6 +145,38 @@ public: ## 其他语言版本 C: +用原来的链表操作： +```c +struct ListNode* removeElements(struct ListNode* head, int val){ + struct ListNode* temp; + // 当头结点存在并且头结点的值等于val时 + while(head && head->val == val) { + temp = head; + // 将新的头结点设置为head->next并删除原来的头结点 + head = head->next; + free(temp); + } + + struct ListNode *cur = head; + // 当cur存在并且cur->next存在时 + // 此解法需要判断cur存在因为cur指向head。若head本身为NULL或者原链表中元素都为val的话，cur也会为NULL + while(cur && (temp = cur->next)) { + // 若cur->next的值等于val + if(temp->val == val) { + // 将cur->next设置为cur->next->next并删除cur->next + cur->next = temp->next; + free(temp); + } + // 若cur->next不等于val，则将cur后移一位 + else + cur = cur->next; + } + + // 返回头结点 + return head; +} +``` +设置一个虚拟头结点： ```c /** * Definition for singly-linked list. From 7cf68bd1f2eaa6ad83d69b969f29b1b146397dfa Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Sun, 24 Apr 2022 17:40:23 +0800 Subject: [PATCH 026/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880738.?= =?UTF-8?q?=E5=8D=95=E8=B0=83=E9=80=92=E5=A2=9E=E7=9A=84=E6=95=B0=E5=AD=97?= =?UTF-8?q?.md=EF=BC=89=EF=BC=9A=E5=A2=9E=E5=8A=A0typescript=E7=89=88?= =?UTF-8?q?=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0738.单调递增的数字.md | 22 ++++++++++++++++++++++ 1 file changed, 22 insertions(+) diff --git a/problems/0738.单调递增的数字.md b/problems/0738.单调递增的数字.md index c8ce8a2b..4e4079a7 100644 --- a/problems/0738.单调递增的数字.md +++ b/problems/0738.单调递增的数字.md @@ -225,6 +225,28 @@ var monotoneIncreasingDigits = function(n) { }; ``` +### TypeScript + +```typescript +function monotoneIncreasingDigits(n: number): number { + let strArr: number[] = String(n).split('').map(i => parseInt(i)); + const length = strArr.length; + let flag: number = length; + for (let i = length - 2; i >= 0; i--) { + if (strArr[i] > strArr[i + 1]) { + strArr[i] -= 1; + flag = i + 1; + } + } + for (let i = flag; i < length; i++) { + strArr[i] = 9; + } + return parseInt(strArr.join('')); +}; +``` + + + -----------------------
From d5f21d534198069fee0dd4a90397d1322d435179 Mon Sep 17 00:00:00 2001 From: eat to 160 pounds <2915390277@qq.com> Date: Sun, 24 Apr 2022 21:05:31 +0800 Subject: [PATCH 027/105] =?UTF-8?q?=E6=9B=B4=E6=96=B0=E4=BA=86=E4=B8=89?= =?UTF-8?q?=E6=95=B0=E4=B9=8B=E5=92=8Cjavascript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0015.三数之和.md | 70 +++++++++++++---------------------- 1 file changed, 26 insertions(+), 44 deletions(-) diff --git a/problems/0015.三数之和.md b/problems/0015.三数之和.md index bfde6b35..cc184c87 100644 --- a/problems/0015.三数之和.md +++ b/problems/0015.三数之和.md @@ -313,54 +313,36 @@ func threeSum(nums []int)[][]int{ javaScript: ```js -/** - * @param {number[]} nums - * @return {number[][]} - */ - -// 循环内不考虑去重 var threeSum = function(nums) { - const len = nums.length; - if(len < 3) return []; - nums.sort((a, b) => a - b); - const resSet = new Set(); - for(let i = 0; i < len - 2; i++) { - if(nums[i] > 0) break; - let l = i + 1, r = len - 1; + const res = [], len = nums.length + // 将数组排序 + nums.sort((a, b) => a - b) + for (let i = 0; i < len; i++) { + let l = i + 1, r = len - 1, iNum = nums[i] + // 数组排过序，如果第一个数大于0直接返回res + if (iNum > 0) return res + // 去重 + if (iNum == nums[i - 1]) continue while(l < r) { - const sum = nums[i] + nums[l] + nums[r]; - if(sum < 0) { l++; continue }; - if(sum > 0) { r--; continue }; - resSet.add(`${nums[i]},${nums[l]},${nums[r]}`); - l++; - r--; + let lNum = nums[l], rNum = nums[r], threeSum = iNum + lNum + rNum + // 三数之和小于0，则左指针向右移动 + if (threeSum < 0) l++ + else if (threeSum > 0) r-- + else { + res.push([iNum, lNum, rNum]) + // 去重 + while(l < r && nums[l] == nums[l + 1]){ + l++ + } + while(l < r && nums[r] == nums[r - 1]) { + r-- + } + l++ + r-- + } } } - return Array.from(resSet).map(i => i.split(",")); -}; - -// 去重优化 -var threeSum = function(nums) { - const len = nums.length; - if(len < 3) return []; - nums.sort((a, b) => a - b); - const res = []; - for(let i = 0; i < len - 2; i++) { - if(nums[i] > 0) break; - // a去重 - if(i > 0 && nums[i] === nums[i - 1]) continue; - let l = i + 1, r = len - 1; - while(l < r) { - const sum = nums[i] + nums[l] + nums[r]; - if(sum < 0) { l++; continue }; - if(sum > 0) { r--; continue }; - res.push([nums[i], nums[l], nums[r]]) - // b c 去重 - while(l < r && nums[l] === nums[++l]); - while(l < r && nums[r] === nums[--r]); - } - } - return res; + return res }; ``` TypeScript: From 84750aca45f96f15e4dc271410c32cde78667e41 Mon Sep 17 00:00:00 2001 From: 243wresfdxvc Date: Sun, 24 Apr 2022 18:32:09 +0000 Subject: [PATCH 028/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200860.=E6=9F=A0?= =?UTF-8?q?=E6=AA=AC=E6=B0=B4=E6=89=BE=E9=9B=B6.md=20C=E8=AF=AD=E8=A8=80?= =?UTF-8?q?=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0860.柠檬水找零.md | 43 ++++++++++++++++++++++++++++++++ 1 file changed, 43 insertions(+) diff --git a/problems/0860.柠檬水找零.md b/problems/0860.柠檬水找零.md index ffd5490d..2738f574 100644 --- a/problems/0860.柠檬水找零.md +++ b/problems/0860.柠檬水找零.md @@ -250,6 +250,49 @@ var lemonadeChange = function(bills) { return true }; +``` +### C +```c +bool lemonadeChange(int* bills, int billsSize){ + // 分别记录五元、十元的数量（二十元不用记录，因为不会用到20元找零） + int fiveCount = 0; int tenCount = 0; + + int i; + for(i = 0; i < billsSize; ++i) { + // 分情况讨论每位顾客的付款 + switch(bills[i]) { + // 情况一：直接收款五元 + case 5: + fiveCount++; + break; + // 情况二：收款十元 + case 10: + // 若没有五元找零，返回false + if(fiveCount == 0) + return false; + // 收款十元并找零五元 + fiveCount--; + tenCount++; + break; + // 情况三：收款二十元 + case 20: + // 若可以，优先用十元和五元找零（因为十元只能找零20，所以需要尽量用掉。而5元能找零十元和二十元） + if(fiveCount > 0 && tenCount > 0) { + fiveCount--; + tenCount--; + } + // 若没有十元，但是有三张五元。用三张五元找零 + else if(fiveCount >= 3) + fiveCount-=3; + // 无法找开，返回false + else + return false; + break; + } + } + // 全部可以找开，返回true + return true; +} ``` ----------------------- From 8291e5e1c6d1b9c913a35015848124ad9cfd1808 Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Mon, 25 Apr 2022 14:05:53 +0800 Subject: [PATCH 029/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880714.?= =?UTF-8?q?=E4=B9=B0=E5=8D=96=E8=82=A1=E7=A5=A8=E7=9A=84=E6=9C=80=E4=BD=B3?= =?UTF-8?q?=E6=97=B6=E6=9C=BA=E5=90=AB=E6=89=8B=E7=BB=AD=E8=B4=B9.md?= =?UTF-8?q?=EF=BC=89=EF=BC=9A=E5=A2=9E=E5=8A=A0typescript=E7=89=88?= =?UTF-8?q?=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- ...买卖股票的最佳时机含手续费.md | 44 +++++++++++++++++++ 1 file changed, 44 insertions(+) diff --git a/problems/0714.买卖股票的最佳时机含手续费.md b/problems/0714.买卖股票的最佳时机含手续费.md index 2f27d6ea..b27631c6 100644 --- a/problems/0714.买卖股票的最佳时机含手续费.md +++ b/problems/0714.买卖股票的最佳时机含手续费.md @@ -293,6 +293,50 @@ var maxProfit = function(prices, fee) { }; ``` +TypeScript： + +> 贪心 + +```typescript +function maxProfit(prices: number[], fee: number): number { + if (prices.length === 0) return 0; + let minPrice: number = prices[0]; + let profit: number = 0; + for (let i = 1, length = prices.length; i < length; i++) { + if (minPrice > prices[i]) { + minPrice = prices[i]; + } + if (minPrice + fee < prices[i]) { + profit += prices[i] - minPrice - fee; + minPrice = prices[i] - fee; + } + } + return profit; +}; +``` + +> 动态规划 + +```typescript +function maxProfit(prices: number[], fee: number): number { + /** + dp[i][1]: 第i天不持有股票的最大所剩现金 + dp[i][0]: 第i天持有股票的最大所剩现金 + */ + const length: number = prices.length; + const dp: number[][] = new Array(length).fill(0).map(_ => []); + dp[0][1] = 0; + dp[0][0] = -prices[0]; + for (let i = 1, length = prices.length; i < length; i++) { + dp[i][1] = Math.max(dp[i - 1][1], dp[i - 1][0] + prices[i] - fee); + dp[i][0] = Math.max(dp[i - 1][0], dp[i - 1][1] - prices[i]); + } + return Math.max(dp[length - 1][0], dp[length - 1][1]); +}; +``` + + + -----------------------
From 7b45a3209014190bbee02de8df566638701db734 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E5=BC=A0=E4=BD=B3=E4=B9=90?= <1791781644@qq.com> Date: Mon, 25 Apr 2022 22:10:18 +0800 Subject: [PATCH 030/105] =?UTF-8?q?=E4=BF=AE=E6=94=B9=20=E7=BB=99=E4=BA=8C?= =?UTF-8?q?=E5=8F=89=E6=A0=91=E9=81=8D=E5=8E=86=E9=A2=98=E7=9B=AE=E5=8A=A0?= =?UTF-8?q?=E4=B8=8A=E5=8A=9B=E6=89=A3=E9=93=BE=E6=8E=A5?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/二叉树的迭代遍历.md | 6 +++--- problems/二叉树的递归遍历.md | 6 +++--- 2 files changed, 6 insertions(+), 6 deletions(-) diff --git a/problems/二叉树的迭代遍历.md b/problems/二叉树的迭代遍历.md index 8164724b..13ba5f1e 100644 --- a/problems/二叉树的迭代遍历.md +++ b/problems/二叉树的迭代遍历.md @@ -11,9 +11,9 @@ 看完本篇大家可以使用迭代法，再重新解决如下三道leetcode上的题目： -* 144.二叉树的前序遍历 -* 94.二叉树的中序遍历 -* 145.二叉树的后序遍历 +* [144.二叉树的前序遍历](https://leetcode-cn.com/problems/binary-tree-preorder-traversal/) +* [94.二叉树的中序遍历](https://leetcode-cn.com/problems/binary-tree-inorder-traversal/) +* [145.二叉树的后序遍历](https://leetcode-cn.com/problems/binary-tree-postorder-traversal/) 为什么可以用迭代法（非递归的方式）来实现二叉树的前后中序遍历呢？ diff --git a/problems/二叉树的递归遍历.md b/problems/二叉树的递归遍历.md index 35d19d7b..13e704ea 100644 --- a/problems/二叉树的递归遍历.md +++ b/problems/二叉树的递归遍历.md @@ -99,9 +99,9 @@ void traversal(TreeNode* cur, vector& vec) { 此时大家可以做一做leetcode上三道题目，分别是： -* 144.二叉树的前序遍历 -* 145.二叉树的后序遍历 -* 94.二叉树的中序遍历 +* [144.二叉树的前序遍历](https://leetcode-cn.com/problems/binary-tree-preorder-traversal/) +* [145.二叉树的后序遍历](https://leetcode-cn.com/problems/binary-tree-postorder-traversal/) +* [94.二叉树的中序遍历](https://leetcode-cn.com/problems/binary-tree-inorder-traversal/) 可能有同学感觉前后中序遍历的递归太简单了，要打迭代法（非递归），别急，我们明天打迭代法，打个通透！ From 9da09584cd39f56c937b5c68b7375ab0d602957b Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Tue, 26 Apr 2022 00:03:54 +0800 Subject: [PATCH 031/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880968.?= =?UTF-8?q?=E7=9B=91=E6=8E=A7=E4=BA=8C=E5=8F=89=E6=A0=91.md=EF=BC=89?= =?UTF-8?q?=EF=BC=9A=E5=A2=9E=E5=8A=A0typescript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0968.监控二叉树.md | 28 ++++++++++++++++++++++++++++ 1 file changed, 28 insertions(+) diff --git a/problems/0968.监控二叉树.md b/problems/0968.监控二叉树.md index 35c3ccdc..9a510a1b 100644 --- a/problems/0968.监控二叉树.md +++ b/problems/0968.监控二叉树.md @@ -476,7 +476,35 @@ var minCameraCover = function(root) { }; ``` +### TypeScript + +```typescript +function minCameraCover(root: TreeNode | null): number { + /** 0-无覆盖， 1-有摄像头， 2-有覆盖 */ + type statusCode = 0 | 1 | 2; + let resCount: number = 0; + if (recur(root) === 0) resCount++; + return resCount; + function recur(node: TreeNode | null): statusCode { + if (node === null) return 2; + const left: statusCode = recur(node.left), + right: statusCode = recur(node.right); + let resStatus: statusCode = 0; + if (left === 0 || right === 0) { + resStatus = 1; + resCount++; + } else if (left === 1 || right === 1) { + resStatus = 2; + } else { + resStatus = 0; + } + return resStatus; + } +}; +``` + ### C + ```c /* **函数后序遍历二叉树。判断一个结点状态时，根据其左右孩子结点的状态进行判断 From 3ec08f41dd08cdd830215bdf2b602bbdcfc92512 Mon Sep 17 00:00:00 2001 From: 243wresfdxvc Date: Mon, 25 Apr 2022 18:18:50 +0000 Subject: [PATCH 032/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200406.=E6=A0=B9?= =?UTF-8?q?=E6=8D=AE=E8=BA=AB=E9=AB=98=E9=87=8D=E5=BB=BA=E9=98=9F=E5=88=97?= =?UTF-8?q?.md=20C=E8=AF=AD=E8=A8=80=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0406.根据身高重建队列.md | 47 +++++++++++++++++++++++ 1 file changed, 47 insertions(+) diff --git a/problems/0406.根据身高重建队列.md b/problems/0406.根据身高重建队列.md index b2354d09..9cefa11c 100644 --- a/problems/0406.根据身高重建队列.md +++ b/problems/0406.根据身高重建队列.md @@ -290,6 +290,53 @@ var reconstructQueue = function(people) { }; ``` +### C +```c +int cmp(const void *p1, const void *p2) { + int *pp1 = *(int**)p1; + int *pp2 = *(int**)p2; + // 若身高相同，则按照k从小到大排列 + // 若身高不同，按身高从大到小排列 + return pp1[0] == pp2[0] ? pp1[1] - pp2[1] : pp2[0] - pp1[0]; +} + +// 将start与end中间的元素都后移一位 +// start为将要新插入元素的位置 +void moveBack(int **people, int peopleSize, int start, int end) { + int i; + for(i = end; i > start; i--) { + people[i] = people[i-1]; + } +} + +int** reconstructQueue(int** people, int peopleSize, int* peopleColSize, int* returnSize, int** returnColumnSizes){ + int i; + // 将people按身高从大到小排列（若身高相同，按k从小到大排列） + qsort(people, peopleSize, sizeof(int*), cmp); + + for(i = 0; i < peopleSize; ++i) { + // people[i]要插入的位置 + int position = people[i][1]; + int *temp = people[i]; + // 将position到i中间的元素后移一位 + // 注：因为已经排好序，position不会比i大。(举例：排序后people最后一位元素最小，其可能的k最大值为peopleSize-2，小于此时的i) + moveBack(people, peopleSize, position, i); + // 将temp放置到position处 + people[position] = temp; + + } + + + // 设置返回二维数组的大小以及里面每个一维数组的长度 + *returnSize = peopleSize; + *returnColumnSizes = (int*)malloc(sizeof(int) * peopleSize); + for(i = 0; i < peopleSize; ++i) { + (*returnColumnSizes)[i] = 2; + } + return people; +} +``` + -----------------------
From c7e34a3f9b82d989b408b4303aee17bd192fb91f Mon Sep 17 00:00:00 2001 From: JackZJ <56966563+laerpeeK@users.noreply.github.com> Date: Thu, 28 Apr 2022 08:28:12 +0800 Subject: [PATCH 033/105] =?UTF-8?q?Update=20=E8=83=8C=E5=8C=85=E7=90=86?= =?UTF-8?q?=E8=AE=BA=E5=9F=BA=E7=A1=8001=E8=83=8C=E5=8C=85JavaScript?= =?UTF-8?q?=E4=BB=A3=E7=A0=81=E5=9D=97?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 更新JavaScript版本二维数组做法，目前文档上的是运行不了的。 --- problems/背包理论基础01背包-1.md | 53 ++++++++++++++---------- 1 file changed, 31 insertions(+), 22 deletions(-) diff --git a/problems/背包理论基础01背包-1.md b/problems/背包理论基础01背包-1.md index fe940b4c..a844dcf5 100644 --- a/problems/背包理论基础01背包-1.md +++ b/problems/背包理论基础01背包-1.md @@ -380,31 +380,40 @@ func main() { ### javascript ```js -function testweightbagproblem (wight, value, size) { - const len = wight.length, - dp = array.from({length: len + 1}).map( - () => array(size + 1).fill(0) - ); - - for(let i = 1; i <= len; i++) { - for(let j = 0; j <= size; j++) { - if(wight[i - 1] <= j) { - dp[i][j] = math.max( - dp[i - 1][j], - value[i - 1] + dp[i - 1][j - wight[i - 1]] - ) - } else { - dp[i][j] = dp[i - 1][j]; - } - } - } +/** + * + * @param {Number []} weight + * @param {Number []} value + * @param {Number} size + * @returns + */ -// console.table(dp); - - return dp[len][size]; +function testWeightBagProblem(weight, value, size) { +const len = weight.length, +dp = Array.from({length: len}).map( +() => Array(size + 1)) //JavaScript 数组是引用类型 +for(let i = 0; i < len; i++) { //初始化最左一列，即背包容量为0时的情况 +dp[i][0] = 0; +} +for(let j = 1; j < size+1; j++) { //初始化第0行, 只有一件物品的情况 +if(weight[0] <= j) { +dp[0][j] = value[0]; +} else { +dp[0][j] = 0; +} +} + +for(let i = 1; i < len; i++) { //dp[i][j]由其左上方元素推导得出 +for(let j = 1; j < size+1; j++) { +if(j < weight[i]) dp[i][j] = dp[i - 1][j]; +else dp[i][j] = Math.max(dp[i-1][j], dp[i-1][j - weight[i]] + value[i]); +} } -function testWeightBagProblem2 (wight, value, size) { +return dp[len-1][size] //满足条件的最大值 +} + +function testWeightBagProblem2 (wight, value, size)4 { const len = wight.length, dp = Array(size + 1).fill(0); for(let i = 1; i <= len; i++) { From 0694fac15bf6b7d18a53adc9f6ad0c33b51db016 Mon Sep 17 00:00:00 2001 From: JackZJ <56966563+laerpeeK@users.noreply.github.com> Date: Thu, 28 Apr 2022 08:30:30 +0800 Subject: [PATCH 034/105] =?UTF-8?q?Update=20=E8=83=8C=E5=8C=85=E7=90=86?= =?UTF-8?q?=E8=AE=BA=E5=9F=BA=E7=A1=8001=E8=83=8C=E5=8C=85JavaScript?= =?UTF-8?q?=E4=BB=A3=E7=A0=81=E5=9D=97?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 更新背包理论基础01背包JavaScript代码块 - 二维数组做法1， 目前主线上的是有问题的。 --- problems/背包理论基础01背包-1.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/背包理论基础01背包-1.md b/problems/背包理论基础01背包-1.md index a844dcf5..d6bc5520 100644 --- a/problems/背包理论基础01背包-1.md +++ b/problems/背包理论基础01背包-1.md @@ -413,7 +413,7 @@ else dp[i][j] = Math.max(dp[i-1][j], dp[i-1][j - weight[i]] + value[i]); return dp[len-1][size] //满足条件的最大值 } -function testWeightBagProblem2 (wight, value, size)4 { +function testWeightBagProblem2 (wight, value, size) { const len = wight.length, dp = Array(size + 1).fill(0); for(let i = 1; i <= len; i++) { From 5fd43cfd0a5770e0cb532687766f64306b5bbfed Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E8=91=A3xx?= Date: Thu, 28 Apr 2022 22:20:22 +0800 Subject: [PATCH 035/105] =?UTF-8?q?0300.=E6=9C=80=E9=95=BF=E4=B8=8A?= =?UTF-8?q?=E5=8D=87=E5=AD=90=E5=BA=8F=E5=88=97-go=E5=8A=A8=E6=80=81?= =?UTF-8?q?=E8=A7=84=E5=88=92=E6=B1=82=E8=A7=A3?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0300.最长上升子序列.md | 33 ++++++++++++++++++++++++++ 1 file changed, 33 insertions(+) diff --git a/problems/0300.最长上升子序列.md b/problems/0300.最长上升子序列.md index dfdd5125..f68edb5a 100644 --- a/problems/0300.最长上升子序列.md +++ b/problems/0300.最长上升子序列.md @@ -168,6 +168,39 @@ func lengthOfLIS(nums []int ) int { } ``` +```go +// 动态规划求解 +func lengthOfLIS(nums []int) int { + // dp数组的定义 dp[i]表示取第i个元素的时候，表示子序列的长度，其中包括 nums[i] 这个元素 + dp := make([]int, len(nums)) + + // 初始化，所有的元素都应该初始化为1 + for i := range dp { + dp[i] = 1 + } + + ans := dp[0] + for i := 1; i < len(nums); i++ { + for j := 0; j < i; j++ { + if nums[i] > nums[j] { + dp[i] = max(dp[i], dp[j] + 1) + } + } + if dp[i] > ans { + ans = dp[i] + } + } + return ans +} + +func max(x, y int) int { + if x > y { + return x + } + return y +} +``` + Javascript ```javascript const lengthOfLIS = (nums) => { From 6f85c2ad3a6daf8e0098b08e61b7fdd68d78c757 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E8=91=A3xx?= Date: Fri, 29 Apr 2022 00:50:22 +0800 Subject: [PATCH 036/105] =?UTF-8?q?0674.=E6=9C=80=E9=95=BF=E8=BF=9E?= =?UTF-8?q?=E7=BB=AD=E9=80=92=E5=A2=9E=E5=BA=8F=E5=88=97-go=E5=8A=A8?= =?UTF-8?q?=E6=80=81=E8=A7=84=E5=88=92=E6=B1=82=E8=A7=A3?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0674.最长连续递增序列.md | 39 +++++++++++++++++++++++ 1 file changed, 39 insertions(+) diff --git a/problems/0674.最长连续递增序列.md b/problems/0674.最长连续递增序列.md index e941d242..56e95d97 100644 --- a/problems/0674.最长连续递增序列.md +++ b/problems/0674.最长连续递增序列.md @@ -236,6 +236,45 @@ class Solution: ``` Go： +> 动态规划： +```go +func findLengthOfLCIS(nums []int) int { + if len(nums) == 0 {return 0} + res, count := 1, 1 + for i := 0; i < len(nums)-1; i++ { + if nums[i+1] > nums[i] { + count++ + }else { + count = 1 + } + if count > res { + res = count + } + } + return res +} +``` + +> 贪心算法： +```go +func findLengthOfLCIS(nums []int) int { + if len(nums) == 0 {return 0} + dp := make([]int, len(nums)) + for i := 0; i < len(dp); i++ { + dp[i] = 1 + } + res := 1 + for i := 0; i < len(nums)-1; i++ { + if nums[i+1] > nums[i] { + dp[i+1] = dp[i] + 1 + } + if dp[i+1] > res { + res = dp[i+1] + } + } + return res +} +``` Javascript： From 31eb619ea18d05e004da853d70b165ba47291c92 Mon Sep 17 00:00:00 2001 From: Anmizi <1845513904@qq.com> Date: Fri, 29 Apr 2022 19:15:41 +0800 Subject: [PATCH 037/105] =?UTF-8?q?=E4=BC=98=E5=8C=96JS=E7=89=88=E6=9C=AC?= =?UTF-8?q?=E4=BB=A3=E7=A0=81=20(0039.=E7=BB=84=E5=90=88=E6=80=BB=E5=92=8C?= =?UTF-8?q?.md)?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0039.组合总和.md | 5 ++--- 1 file changed, 2 insertions(+), 3 deletions(-) diff --git a/problems/0039.组合总和.md b/problems/0039.组合总和.md index 98b37b84..e10a827f 100644 --- a/problems/0039.组合总和.md +++ b/problems/0039.组合总和.md @@ -370,18 +370,17 @@ func backtracking(startIndex,sum,target int,candidates,trcak []int,res *[][]int) ```js var combinationSum = function(candidates, target) { const res = [], path = []; - candidates.sort(); // 排序 + candidates.sort((a,b)=>a-b); // 排序 backtracking(0, 0); return res; function backtracking(j, sum) { - if (sum > target) return; if (sum === target) { res.push(Array.from(path)); return; } for(let i = j; i < candidates.length; i++ ) { const n = candidates[i]; - if(n > target - sum) continue; + if(n > target - sum) break; path.push(n); sum += n; backtracking(i, sum); From dac0b4a12e43670eb54eaa6da5fa3f07cb707424 Mon Sep 17 00:00:00 2001 From: GitHubQAQ <31883473+GitHubQAQ@users.noreply.github.com> Date: Fri, 29 Apr 2022 21:03:27 +0800 Subject: [PATCH 038/105] =?UTF-8?q?Update=200106.=E4=BB=8E=E4=B8=AD?= =?UTF-8?q?=E5=BA=8F=E4=B8=8E=E5=90=8E=E5=BA=8F=E9=81=8D=E5=8E=86=E5=BA=8F?= =?UTF-8?q?=E5=88=97=E6=9E=84=E9=80=A0=E4=BA=8C=E5=8F=89=E6=A0=91.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 优化代码高亮 --- .../0106.从中序与后序遍历序列构造二叉树.md | 6 +++--- 1 file changed, 3 insertions(+), 3 deletions(-) diff --git a/problems/0106.从中序与后序遍历序列构造二叉树.md b/problems/0106.从中序与后序遍历序列构造二叉树.md index 496de431..4396bc76 100644 --- a/problems/0106.从中序与后序遍历序列构造二叉树.md +++ b/problems/0106.从中序与后序遍历序列构造二叉树.md @@ -103,7 +103,7 @@ TreeNode* traversal (vector& inorder, vector& postorder) { 中序数组相对比较好切，找到切割点（后序数组的最后一个元素）在中序数组的位置，然后切割，如下代码中我坚持左闭右开的原则： -```C++ +```CPP // 找到中序遍历的切割点 int delimiterIndex; for (delimiterIndex = 0; delimiterIndex < inorder.size(); delimiterIndex++) { @@ -130,7 +130,7 @@ vector rightInorder(inorder.begin() + delimiterIndex + 1, inorder.end() ); 代码如下： -``` +```CPP // postorder 舍弃末尾元素，因为这个元素就是中间节点，已经用过了 postorder.resize(postorder.size() - 1); @@ -144,7 +144,7 @@ vector rightPostorder(postorder.begin() + leftInorder.size(), postorder.end 接下来可以递归了，代码如下： -``` +```CPP root->left = traversal(leftInorder, leftPostorder); root->right = traversal(rightInorder, rightPostorder); ``` From 5b3607c6a151b6fd6dd548c1b26f3f1fb8514047 Mon Sep 17 00:00:00 2001 From: Anmizi <1845513904@qq.com> Date: Fri, 29 Apr 2022 23:18:33 +0800 Subject: [PATCH 039/105] =?UTF-8?q?=E4=BF=AE=E6=94=B9=200040.=E7=BB=84?= =?UTF-8?q?=E5=90=88=E6=80=BB=E5=92=8CII.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0040.组合总和II.md | 15 ++++++++++----- 1 file changed, 10 insertions(+), 5 deletions(-) diff --git a/problems/0040.组合总和II.md b/problems/0040.组合总和II.md index de13e031..34ac64e6 100644 --- a/problems/0040.组合总和II.md +++ b/problems/0040.组合总和II.md @@ -508,22 +508,27 @@ func backtracking(startIndex,sum,target int,candidates,trcak []int,res *[][]int) */ var combinationSum2 = function(candidates, target) { const res = []; path = [], len = candidates.length; - candidates.sort(); + candidates.sort((a,b)=>a-b); backtracking(0, 0); return res; function backtracking(sum, i) { - if (sum > target) return; if (sum === target) { res.push(Array.from(path)); return; } - let f = -1; for(let j = i; j < len; j++) { const n = candidates[j]; - if(n > target - sum || n === f) continue; + if(j > i && candidates[j] === candidates[j-1]){ + //若当前元素和前一个元素相等 + //则本次循环结束，防止出现重复组合 + continue; + } + //如果当前元素值大于目标值-总和的值 + //由于数组已排序，那么该元素之后的元素必定不满足条件 + //直接终止当前层的递归 + if(n > target - sum) break; path.push(n); sum += n; - f = n; backtracking(sum, j + 1); path.pop(); sum -= n; From 55c78be1280054615f23c9e5859d139a7f840705 Mon Sep 17 00:00:00 2001 From: dmzlingyin Date: Fri, 29 Apr 2022 23:38:34 +0800 Subject: [PATCH 040/105] =?UTF-8?q?update=20(0739.=E6=AF=8F=E6=97=A5?= =?UTF-8?q?=E6=B8=A9=E5=BA=A6.md):=20python=E4=BB=A3=E7=A0=81=E9=AB=98?= =?UTF-8?q?=E4=BA=AE?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0739.每日温度.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0739.每日温度.md b/problems/0739.每日温度.md index 710f5eb6..7deab0a3 100644 --- a/problems/0739.每日温度.md +++ b/problems/0739.每日温度.md @@ -233,7 +233,7 @@ class Solution { } ``` Python： -``` Python3 +```python class Solution: def dailyTemperatures(self, temperatures: List[int]) -> List[int]: answer = [0]*len(temperatures) From d0a79760b65a1078284b036be6dcf8fa8eec5076 Mon Sep 17 00:00:00 2001 From: dmzlingyin Date: Fri, 29 Apr 2022 23:41:26 +0800 Subject: [PATCH 041/105] =?UTF-8?q?update=20(0739.=E6=AF=8F=E6=97=A5?= =?UTF-8?q?=E6=B8=A9=E5=BA=A6.md):=20=E8=AF=AD=E8=A8=80=E8=A1=A8=E8=BF=B0?= =?UTF-8?q?=E4=BF=AE=E5=A4=8D?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0739.每日温度.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0739.每日温度.md b/problems/0739.每日温度.md index 7deab0a3..f0f782d2 100644 --- a/problems/0739.每日温度.md +++ b/problems/0739.每日温度.md @@ -34,7 +34,7 @@ 那么单调栈的原理是什么呢？为什么时间复杂度是O(n)就可以找到每一个元素的右边第一个比它大的元素位置呢？ -单调栈的本质是空间换时间，因为在遍历的过程中需要用一个栈来记录右边第一个比当前元素的元素，优点是只需要遍历一次。 +单调栈的本质是空间换时间，因为在遍历的过程中需要用一个栈来记录右边第一个比当前元素大的元素，优点是只需要遍历一次。 在使用单调栈的时候首先要明确如下几点： From 636550d44c4ac65521865cca00b1c3c8b9f7c756 Mon Sep 17 00:00:00 2001 From: dmzlingyin Date: Fri, 29 Apr 2022 23:50:31 +0800 Subject: [PATCH 042/105] =?UTF-8?q?update=20(0739.=E6=AF=8F=E6=97=A5?= =?UTF-8?q?=E6=B8=A9=E5=BA=A6.md):=20=E5=A2=9E=E5=8A=A0=E6=9C=AA=E7=B2=BE?= =?UTF-8?q?=E7=AE=80=E7=89=88=E6=9C=ACGo=E4=BB=A3=E7=A0=81?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0739.每日温度.md | 30 +++++++++++++++++++++++++++++- 1 file changed, 29 insertions(+), 1 deletion(-) diff --git a/problems/0739.每日温度.md b/problems/0739.每日温度.md index f0f782d2..206bebd2 100644 --- a/problems/0739.每日温度.md +++ b/problems/0739.每日温度.md @@ -277,8 +277,36 @@ func dailyTemperatures(t []int) []int { } ``` -> 单调栈法 +> 单调栈法(未精简版本) +```go +func dailyTemperatures(temperatures []int) []int { + res := make([]int, len(temperatures)) + // 初始化栈顶元素为第一个下标索引0 + stack := []int{0} + + for i := 1; i < len(temperatures); i++ { + top := stack[len(stack)-1] + if temperatures[i] < temperatures[top] { + stack = append(stack, i) + } else if temperatures[i] == temperatures[top] { + stack = append(stack, i) + } else { + for len(stack) != 0 && temperatures[i] > temperatures[top] { + res[top] = i - top + stack = stack[:len(stack)-1] + if len(stack) != 0 { + top = stack[len(stack)-1] + } + } + stack = append(stack, i) + } + } + return res +} +``` + +> 单调栈法(精简版本) ```go // 单调递减栈 func dailyTemperatures(num []int) []int { From 49ec574821f687ea8703c61245f815bf2dc76139 Mon Sep 17 00:00:00 2001 From: dmzlingyin Date: Sat, 30 Apr 2022 11:04:17 +0800 Subject: [PATCH 043/105] =?UTF-8?q?update=20(0496.=E4=B8=8B=E4=B8=80?= =?UTF-8?q?=E4=B8=AA=E6=9B=B4=E5=A4=A7=E5=85=83=E7=B4=A0I):=20=E5=A2=9E?= =?UTF-8?q?=E5=8A=A0=E6=9C=AA=E7=B2=BE=E7=AE=80Go=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0496.下一个更大元素I.md | 33 +++++++++++++++++++++++++ 1 file changed, 33 insertions(+) diff --git a/problems/0496.下一个更大元素I.md b/problems/0496.下一个更大元素I.md index f9dfa308..02339677 100644 --- a/problems/0496.下一个更大元素I.md +++ b/problems/0496.下一个更大元素I.md @@ -244,6 +244,39 @@ class Solution: ``` Go： + +> 未精简版本 +```go +func nextGreaterElement(nums1 []int, nums2 []int) []int { + res := make([]int, len(nums1)) + for i := range res { res[i] = -1 } + m := make(map[int]int, len(nums1)) + for k, v := range nums1 { m[v] = k } + + stack := []int{0} + for i := 1; i < len(nums2); i++ { + top := stack[len(stack)-1] + if nums2[i] < nums2[top] { + stack = append(stack, i) + } else if nums2[i] == nums2[top] { + stack = append(stack, i) + } else { + for len(stack) != 0 && nums2[i] > nums2[top] { + if v, ok := m[nums2[top]]; ok { + res[v] = nums2[i] + } + stack = stack[:len(stack)-1] + if len(stack) != 0 { + top = stack[len(stack)-1] + } + } + stack = append(stack, i) + } + } + return res +} +``` +> 精简版本 ```go func nextGreaterElement(nums1 []int, nums2 []int) []int { res := make([]int, len(nums1)) From 82e8b9eaac23041c2a358af126ec4e268ee383fa Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Sat, 30 Apr 2022 21:18:55 +0800 Subject: [PATCH 044/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880509.?= =?UTF-8?q?=E6=96=90=E6=B3=A2=E9=82=A3=E5=A5=91=E6=95=B0.md=EF=BC=89?= =?UTF-8?q?=EF=BC=9A=E5=A2=9E=E5=8A=A0typescript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0509.斐波那契数.md | 22 ++++++++++++++++++++++ 1 file changed, 22 insertions(+) diff --git a/problems/0509.斐波那契数.md b/problems/0509.斐波那契数.md index d339940c..1d17784d 100644 --- a/problems/0509.斐波那契数.md +++ b/problems/0509.斐波那契数.md @@ -245,7 +245,29 @@ var fib = function(n) { }; ``` +TypeScript + +```typescript +function fib(n: number): number { + /** + dp[i]: 第i个斐波那契数 + dp[0]: 0; + dp[1]：1； + ... + dp[i] = dp[i - 1] + dp[i - 2]; + */ + const dp: number[] = []; + dp[0] = 0; + dp[1] = 1; + for (let i = 2; i <= n; i++) { + dp[i] = dp[i - 1] + dp[i - 2]; + } + return dp[n]; +}; +``` + ### C + 动态规划: ```c int fib(int n){ From 0cba2d22f2d032bbf478662ef622d8022af24f21 Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Sat, 30 Apr 2022 21:53:14 +0800 Subject: [PATCH 045/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880070.?= =?UTF-8?q?=E7=88=AC=E6=A5=BC=E6=A2=AF.md=EF=BC=89=EF=BC=9A=E5=A2=9E?= =?UTF-8?q?=E5=8A=A0typescript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0070.爬楼梯.md | 51 ++++++++++++++++++++++++++++++++++++++ 1 file changed, 51 insertions(+) diff --git a/problems/0070.爬楼梯.md b/problems/0070.爬楼梯.md index da19ea0e..34d41441 100644 --- a/problems/0070.爬楼梯.md +++ b/problems/0070.爬楼梯.md @@ -308,7 +308,58 @@ var climbStairs = function(n) { }; ``` +TypeScript + +> 爬2阶 + +```typescript +function climbStairs(n: number): number { + /** + dp[i]: i阶楼梯的方法种数 + dp[1]: 1; + dp[2]: 2; + ... + dp[i]: dp[i - 1] + dp[i - 2]; + */ + const dp: number[] = []; + dp[1] = 1; + dp[2] = 2; + for (let i = 3; i <= n; i++) { + dp[i] = dp[i - 1] + dp[i - 2]; + } + return dp[n]; +}; +``` + +> 爬m阶 + +```typescript +function climbStairs(n: number): number { + /** + 一次可以爬m阶 + dp[i]: i阶楼梯的方法种数 + dp[1]: 1; + dp[2]: 2; + dp[3]: dp[2] + dp[1]; + ... + dp[i]: dp[i - 1] + dp[i - 2] + ... + dp[max(i - m, 1)]; 从i-1加到max(i-m, 1) + */ + const m: number = 2; // 本题m为2 + const dp: number[] = new Array(n + 1).fill(0); + dp[1] = 1; + dp[2] = 2; + for (let i = 3; i <= n; i++) { + const end: number = Math.max(i - m, 1); + for (let j = i - 1; j >= end; j--) { + dp[i] += dp[j]; + } + } + return dp[n]; +}; +``` + ### C + ```c int climbStairs(int n){ //若n<=2，返回n From 1e81ef27100c8fc76bc38ec0135a4b18554398af Mon Sep 17 00:00:00 2001 From: Beim <73528776+162-jld@users.noreply.github.com> Date: Sun, 1 May 2022 11:51:11 +0800 Subject: [PATCH 046/105] =?UTF-8?q?Update=200383.=E8=B5=8E=E9=87=91?= =?UTF-8?q?=E4=BF=A1.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0383.赎金信.md | 28 +++++++++++++++------------- 1 file changed, 15 insertions(+), 13 deletions(-) diff --git a/problems/0383.赎金信.md b/problems/0383.赎金信.md index 00707347..56dcb8dd 100644 --- a/problems/0383.赎金信.md +++ b/problems/0383.赎金信.md @@ -110,23 +110,25 @@ Java： ```Java class Solution { public boolean canConstruct(String ransomNote, String magazine) { - //记录杂志字符串出现的次数 - int[] arr = new int[26]; - int temp; - for (int i = 0; i < magazine.length(); i++) { - temp = magazine.charAt(i) - 'a'; - arr[temp]++; + // 定义一个哈希映射数组 + int[] record = new int[26]; + + // 遍历 + for(char c : magazine.toCharArray()){ + record[c - 'a'] += 1; } - for (int i = 0; i < ransomNote.length(); i++) { - temp = ransomNote.charAt(i) - 'a'; - //对于金信中的每一个字符都在数组中查找 - //找到相应位减一，否则找不到返回false - if (arr[temp] > 0) { - arr[temp]--; - } else { + + for(char c : ransomNote.toCharArray()){ + record[c - 'a'] -= 1; + } + + // 如果数组中存在负数，说明ransomNote字符串总存在magazine中没有的字符 + for(int i : record){ + if(i < 0){ return false; } } + return true; } } From 25e26f1f86bf246f93560710f41dcf5a26414770 Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Sun, 1 May 2022 13:45:25 +0800 Subject: [PATCH 047/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0(0746.=E4=BD=BF?= =?UTF-8?q?=E7=94=A8=E6=9C=80=E5=B0=8F=E8=8A=B1=E8=B4=B9=E7=88=AC=E6=A5=BC?= =?UTF-8?q?=E6=A2=AF.md)=EF=BC=9A=E5=A2=9E=E5=8A=A0typescript=E7=89=88?= =?UTF-8?q?=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0746.使用最小花费爬楼梯.md | 23 ++++++++++++++++++++ 1 file changed, 23 insertions(+) diff --git a/problems/0746.使用最小花费爬楼梯.md b/problems/0746.使用最小花费爬楼梯.md index c356955a..5931fc8a 100644 --- a/problems/0746.使用最小花费爬楼梯.md +++ b/problems/0746.使用最小花费爬楼梯.md @@ -266,7 +266,30 @@ var minCostClimbingStairs = function(cost) { }; ``` +### TypeScript + +```typescript +function minCostClimbingStairs(cost: number[]): number { + /** + dp[i]: 走到第i阶需要花费的最少金钱 + dp[0]: cost[0]; + dp[1]: cost[1]; + ... + dp[i]: min(dp[i - 1], dp[i - 2]) + cost[i]; + */ + const dp: number[] = []; + const length: number = cost.length; + dp[0] = cost[0]; + dp[1] = cost[1]; + for (let i = 2; i <= length; i++) { + dp[i] = Math.min(dp[i - 1], dp[i - 2]) + cost[i]; + } + return Math.min(dp[length - 1], dp[length - 2]); +}; +``` + ### C + ```c int minCostClimbingStairs(int* cost, int costSize){ //开辟dp数组，大小为costSize From fc28660b6189b010709bedc7dd5acf678eaf47ed Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Sun, 1 May 2022 14:42:12 +0800 Subject: [PATCH 048/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880062.?= =?UTF-8?q?=E4=B8=8D=E5=90=8C=E8=B7=AF=E5=BE=84.md=EF=BC=89=EF=BC=9A?= =?UTF-8?q?=E5=A2=9E=E5=8A=A0typescript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0062.不同路径.md | 30 +++++++++++++++++++++++++++++- 1 file changed, 29 insertions(+), 1 deletion(-) diff --git a/problems/0062.不同路径.md b/problems/0062.不同路径.md index 4a9af129..f59b7be8 100644 --- a/problems/0062.不同路径.md +++ b/problems/0062.不同路径.md @@ -273,7 +273,7 @@ public: return dp[m-1][n-1]; } -``` +``` ### Python ```python @@ -347,7 +347,35 @@ var uniquePaths = function(m, n) { }; ``` +### TypeScript + +```typescript +function uniquePaths(m: number, n: number): number { + /** + dp[i][j]: 到达(i, j)的路径数 + dp[0][*]: 1; + dp[*][0]: 1; + ... + dp[i][j]: dp[i - 1][j] + dp[i][j - 1]; + */ + const dp: number[][] = new Array(m).fill(0).map(_ => []); + for (let i = 0; i < m; i++) { + dp[i][0] = 1; + } + for (let i = 0; i < n; i++) { + dp[0][i] = 1; + } + for (let i = 1; i < m; i++) { + for (let j = 1; j < n; j++) { + dp[i][j] = dp[i - 1][j] + dp[i][j - 1]; + } + } + return dp[m - 1][n - 1]; +}; +``` + ### C + ```c //初始化dp数组 int **initDP(int m, int n) { From 8c596b161a140a76ea57130ac61a009308bd277c Mon Sep 17 00:00:00 2001 From: Beim <1497359184@qq.com> Date: Sun, 1 May 2022 15:45:58 +0800 Subject: [PATCH 049/105] modify_problems_0383 --- README.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/README.md b/README.md index 1d7f219d..3d44ca69 100644 --- a/README.md +++ b/README.md @@ -31,7 +31,7 @@

-# LeetCode 刷题攻略 +# LeetCode 刷题攻略1111 ## 刷题攻略的背景 @@ -254,7 +254,7 @@ 33. [二叉树：构造一棵搜索树](./problems/0108.将有序数组转换为二叉搜索树.md) 34. [二叉树：搜索树转成累加树](./problems/0538.把二叉搜索树转换为累加树.md) 35. [二叉树：总结篇！（需要掌握的二叉树技能都在这里了）](./problems/二叉树总结篇.md) - + ## 回溯算法 题目分类大纲如下： From d3e1f1d3b3406f28e7494340928fbc1866c0b6f6 Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Sun, 1 May 2022 19:24:05 +0800 Subject: [PATCH 050/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880063.?= =?UTF-8?q?=E4=B8=8D=E5=90=8C=E8=B7=AF=E5=BE=84II.md=EF=BC=89=EF=BC=9A?= =?UTF-8?q?=E5=A2=9E=E5=8A=A0typescript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0063.不同路径II.md | 33 ++++++++++++++++++++++++++++++++- 1 file changed, 32 insertions(+), 1 deletion(-) diff --git a/problems/0063.不同路径II.md b/problems/0063.不同路径II.md index a40cceda..d09ea0e6 100644 --- a/problems/0063.不同路径II.md +++ b/problems/0063.不同路径II.md @@ -352,7 +352,38 @@ var uniquePathsWithObstacles = function(obstacleGrid) { }; ``` -C +### TypeScript + +```typescript +function uniquePathsWithObstacles(obstacleGrid: number[][]): number { + /** + dp[i][j]: 到达(i, j)的路径数 + dp[0][*]: 用u表示第一个障碍物下标，则u之前为1，u之后（含u）为0 + dp[*][0]: 同上 + ... + dp[i][j]: obstacleGrid[i][j] === 1 ? 0 : dp[i-1][j] + dp[i][j-1]; + */ + const m: number = obstacleGrid.length; + const n: number = obstacleGrid[0].length; + const dp: number[][] = new Array(m).fill(0).map(_ => new Array(n).fill(0)); + for (let i = 0; i < m && obstacleGrid[i][0] === 0; i++) { + dp[i][0] = 1; + } + for (let i = 0; i < n && obstacleGrid[0][i] === 0; i++) { + dp[0][i] = 1; + } + for (let i = 1; i < m; i++) { + for (let j = 1; j < n; j++) { + if (obstacleGrid[i][j] === 1) continue; + dp[i][j] = dp[i - 1][j] + dp[i][j - 1]; + } + } + return dp[m - 1][n - 1]; +}; +``` + +### C + ```c //初始化dp数组 int **initDP(int m, int n, int** obstacleGrid) { From dd514eb08787dd203cea4668f75477971b4e88fe Mon Sep 17 00:00:00 2001 From: wang <472146630@qq.com> Date: Sun, 1 May 2022 21:08:57 +0800 Subject: [PATCH 051/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200053.=E6=9C=80?= =?UTF-8?q?=E5=A4=A7=E5=AD=90=E5=BA=8F=E5=92=8C=E3=80=810135=E5=88=86?= =?UTF-8?q?=E5=8F=91=E7=B3=96=E6=9E=9C=E5=92=8C0455=E5=88=86=E5=8F=91?= =?UTF-8?q?=E9=A5=BC=E5=B9=B2=E7=9A=84=20Rust=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0053.最大子序和.md | 20 +++++++++++++++--- problems/0135.分发糖果.md | 29 ++++++++++++++++++++----- problems/0455.分发饼干.md | 36 +++++++++++++++++++++++++------- 3 files changed, 69 insertions(+), 16 deletions(-) diff --git a/problems/0053.最大子序和.md b/problems/0053.最大子序和.md index b5fb7642..9dbc7313 100644 --- a/problems/0053.最大子序和.md +++ b/problems/0053.最大子序和.md @@ -140,7 +140,7 @@ public: ## 其他语言版本 -### Java +### Java ```java class Solution { public int maxSubArray(int[] nums) { @@ -180,7 +180,7 @@ class Solution { } ``` -### Python +### Python ```python class Solution: def maxSubArray(self, nums: List[int]) -> int: @@ -195,7 +195,7 @@ class Solution: return result ``` -### Go +### Go ```go func maxSubArray(nums []int) int { @@ -212,6 +212,20 @@ func maxSubArray(nums []int) int { } ``` +### Rust +```rust +pub fn max_sub_array(nums: Vec) -> i32 { + let mut max_sum = i32::MIN; + let mut curr = 0; + for n in nums.iter() { + curr += n; + max_sum = max_sum.max(curr); + curr = curr.max(0); + } + max_sum +} +``` + ### Javascript: ```Javascript var maxSubArray = function(nums) { diff --git a/problems/0135.分发糖果.md b/problems/0135.分发糖果.md index ccdabc16..ce738689 100644 --- a/problems/0135.分发糖果.md +++ b/problems/0135.分发糖果.md @@ -126,11 +126,11 @@ public: ## 其他语言版本 -### Java +### Java ```java class Solution { - /** - 分两个阶段 + /** + 分两个阶段 1、起点下标1 从左往右，只要 右边 比 左边 大，右边的糖果=左边 + 1 2、起点下标 ratings.length - 2 从右往左， 只要左边 比 右边 大，此时 左边的糖果应该 取本身的糖果数（符合比它左边大） 和 右边糖果数 + 1 二者的最大值，这样才符合 它比它左边的大，也比它右边大 */ @@ -160,7 +160,7 @@ class Solution { } ``` -### Python +### Python ```python class Solution: def candy(self, ratings: List[int]) -> int: @@ -213,6 +213,25 @@ func findMax(num1 int ,num2 int) int{ } ``` +### Rust +```rust +pub fn candy(ratings: Vec) -> i32 { + let mut candies = vec![1i32; ratings.len()]; + for i in 1..ratings.len() { + if ratings[i - 1] < ratings[i] { + candies[i] = candies[i - 1] + 1; + } + } + + for i in (0..ratings.len()-1).rev() { + if ratings[i] > ratings[i + 1] { + candies[i] = candies[i].max(candies[i + 1] + 1); + } + } + candies.iter().sum() +} +``` + ### Javascript: ```Javascript var candy = function(ratings) { @@ -229,7 +248,7 @@ var candy = function(ratings) { candys[i] = Math.max(candys[i], candys[i + 1] + 1) } } - + let count = candys.reduce((a, b) => { return a + b }) diff --git a/problems/0455.分发饼干.md b/problems/0455.分发饼干.md index d95a407a..17db4a85 100644 --- a/problems/0455.分发饼干.md +++ b/problems/0455.分发饼干.md @@ -106,7 +106,7 @@ public: ## 其他语言版本 -### Java +### Java ```java class Solution { // 思路1：优先考虑饼干，小饼干先喂饱小胃口 @@ -145,7 +145,7 @@ class Solution { } ``` -### Python +### Python ```python class Solution: # 思路1：优先考虑胃饼干 @@ -166,13 +166,13 @@ class Solution: s.sort() start, count = len(s) - 1, 0 for index in range(len(g) - 1, -1, -1): # 先喂饱大胃口 - if start >= 0 and g[index] <= s[start]: + if start >= 0 and g[index] <= s[start]: start -= 1 count += 1 return count ``` -### Go +### Go ```golang //排序后，局部最优 func findContentChildren(g []int, s []int) int { @@ -191,7 +191,27 @@ func findContentChildren(g []int, s []int) int { } ``` -### Javascript +### Rust +```rust +pub fn find_content_children(children: Vec, cookie: Vec) -> i32 { + let mut children = children; + let mut cookies = cookie; + children.sort(); + cookies.sort(); + + let (mut child, mut cookie) = (0usize, 0usize); + while child < children.len() && cookie < cookies.len() { + // 优先选择最小饼干喂饱孩子 + if children[child] <= cookies[cookie] { + child += 1; + } + cookie += 1 + } + child as i32 +} +``` + +### Javascript ```js var findContentChildren = function(g, s) { g = g.sort((a, b) => a - b) @@ -203,7 +223,7 @@ var findContentChildren = function(g, s) { result++ index-- } - } + } return result }; @@ -251,7 +271,7 @@ function findContentChildren(g: number[], s: number[]): number { }; ``` -### C +### C ```c int cmp(int* a, int* b) { @@ -261,7 +281,7 @@ int cmp(int* a, int* b) { int findContentChildren(int* g, int gSize, int* s, int sSize){ if(sSize == 0) return 0; - + //将两个数组排序为升序 qsort(g, gSize, sizeof(int), cmp); qsort(s, sSize, sizeof(int), cmp); From 4c03ad7a78fcb324da45e97a2cfca618af1086c8 Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Sun, 1 May 2022 22:32:12 +0800 Subject: [PATCH 052/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880343.?= =?UTF-8?q?=E6=95=B4=E6=95=B0=E6=8B=86=E5=88=86.md=EF=BC=89=EF=BC=9A?= =?UTF-8?q?=E5=A2=9E=E5=8A=A0typescript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0343.整数拆分.md | 28 +++++++++++++++++++++++++++- 1 file changed, 27 insertions(+), 1 deletion(-) diff --git a/problems/0343.整数拆分.md b/problems/0343.整数拆分.md index 4a7ba6ab..279f1d71 100644 --- a/problems/0343.整数拆分.md +++ b/problems/0343.整数拆分.md @@ -274,7 +274,33 @@ var integerBreak = function(n) { }; ``` -C: +### TypeScript + +```typescript +function integerBreak(n: number): number { + /** + dp[i]: i对应的最大乘积 + dp[2]: 1; + ... + dp[i]: max( + 1 * dp[i - 1], 1 * (i - 1), + 2 * dp[i - 2], 2 * (i - 2), + ..., (i - 2) * dp[2], (i - 2) * 2 + ); + */ + const dp: number[] = new Array(n + 1).fill(0); + dp[2] = 1; + for (let i = 3; i <= n; i++) { + for (let j = 1; j <= i - 2; j++) { + dp[i] = Math.max(dp[i], j * dp[i - j], j * (i - j)); + } + } + return dp[n]; +}; +``` + +### C + ```c //初始化DP数组 int *initDP(int num) { From b1ef364ffb9c0082f401c34bdd6bf22c4f330fa5 Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Mon, 2 May 2022 17:25:45 +0800 Subject: [PATCH 053/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880096.?= =?UTF-8?q?=E4=B8=8D=E5=90=8C=E7=9A=84=E4=BA=8C=E5=8F=89=E6=90=9C=E7=B4=A2?= =?UTF-8?q?=E6=A0=91.md=EF=BC=89=EF=BC=9A=E5=A2=9E=E5=8A=A0typescript?= =?UTF-8?q?=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0096.不同的二叉搜索树.md | 28 ++++++++++++++++++++++- 1 file changed, 27 insertions(+), 1 deletion(-) diff --git a/problems/0096.不同的二叉搜索树.md b/problems/0096.不同的二叉搜索树.md index 41fcb8fe..25561b50 100644 --- a/problems/0096.不同的二叉搜索树.md +++ b/problems/0096.不同的二叉搜索树.md @@ -227,7 +227,33 @@ const numTrees =(n) => { }; ``` -C: +TypeScript + +```typescript +function numTrees(n: number): number { + /** + dp[i]: i个节点对应的种树 + dp[0]: -1; 无意义； + dp[1]: 1; + ... + dp[i]: 2 * dp[i - 1] + + (dp[1] * dp[i - 2] + dp[2] * dp[i - 3] + ... + dp[i - 2] * dp[1]); 从1加到i-2 + */ + const dp: number[] = []; + dp[0] = -1; // 表示无意义 + dp[1] = 1; + for (let i = 2; i <= n; i++) { + dp[i] = 2 * dp[i - 1]; + for (let j = 1, end = i - 1; j < end; j++) { + dp[i] += dp[j] * dp[end - j]; + } + } + return dp[n]; +}; +``` + +### C + ```c //开辟dp数组 int *initDP(int n) { From 5440c3d46b16565a265e30e498662829bf40b5e1 Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Tue, 3 May 2022 11:48:03 +0800 Subject: [PATCH 054/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=88=E8=83=8C?= =?UTF-8?q?=E5=8C=85=E7=90=86=E8=AE=BA=E5=9F=BA=E7=A1=8001=E8=83=8C?= =?UTF-8?q?=E5=8C=85-1.md=EF=BC=89=EF=BC=9A=E5=A2=9E=E5=8A=A0typescript?= =?UTF-8?q?=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/背包理论基础01背包-1.md | 42 ++++++++++++++++++++++++ 1 file changed, 42 insertions(+) diff --git a/problems/背包理论基础01背包-1.md b/problems/背包理论基础01背包-1.md index fe940b4c..257e87e4 100644 --- a/problems/背包理论基础01背包-1.md +++ b/problems/背包理论基础01背包-1.md @@ -423,5 +423,47 @@ function test () { test(); ``` +### TypeScript + +```typescript +function testWeightBagProblem( + weight: number[], + value: number[], + size: number +): number { + /** + * dp[i][j]: 前i个物品，背包容量为j，能获得的最大价值 + * dp[0][*]: u=weight[0],u之前为0，u之后（含u）为value[0] + * dp[*][0]: 0 + * ... + * dp[i][j]: max(dp[i-1][j], dp[i-1][j-weight[i]]+value[i]); + */ + const goodsNum: number = weight.length; + const dp: number[][] = new Array(goodsNum) + .fill(0) + .map((_) => new Array(size + 1).fill(0)); + for (let i = weight[0]; i <= size; i++) { + dp[0][i] = value[0]; + } + for (let i = 1; i < goodsNum; i++) { + for (let j = 1; j <= size; j++) { + if (j < weight[i]) { + dp[i][j] = dp[i - 1][j]; + } else { + dp[i][j] = Math.max(dp[i - 1][j], dp[i - 1][j - weight[i]] + value[i]); + } + } + } + return dp[goodsNum - 1][size]; +} +// test +const weight = [1, 3, 4]; +const value = [15, 20, 30]; +const size = 4; +console.log(testWeightBagProblem(weight, value, size)); +``` + + + -----------------------
From 36f630b109a93685bb241884e5584b327028d5f5 Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Tue, 3 May 2022 12:34:20 +0800 Subject: [PATCH 055/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=88=E8=83=8C?= =?UTF-8?q?=E5=8C=85=E7=90=86=E8=AE=BA=E5=9F=BA=E7=A1=8001=E8=83=8C?= =?UTF-8?q?=E5=8C=85-2.md=EF=BC=89=EF=BC=9A=E5=A2=9E=E5=8A=A0typescript?= =?UTF-8?q?=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/背包理论基础01背包-2.md | 24 ++++++++++++++++++++++++ 1 file changed, 24 insertions(+) diff --git a/problems/背包理论基础01背包-2.md b/problems/背包理论基础01背包-2.md index dabdfb2d..bb5e3a03 100644 --- a/problems/背包理论基础01背包-2.md +++ b/problems/背包理论基础01背包-2.md @@ -315,6 +315,30 @@ function test () { test(); ``` +### TypeScript + +```typescript +function testWeightBagProblem( + weight: number[], + value: number[], + size: number +): number { + const goodsNum: number = weight.length; + const dp: number[] = new Array(size + 1).fill(0); + for (let i = 0; i < goodsNum; i++) { + for (let j = size; j >= weight[i]; j--) { + dp[j] = Math.max(dp[j], dp[j - weight[i]] + value[i]); + } + } + return dp[size]; +} +const weight = [1, 3, 4]; +const value = [15, 20, 30]; +const size = 4; +console.log(testWeightBagProblem(weight, value, size)); + +``` + ----------------------- From a38ee5f525af998683a2d0fb38f6c6cac1312f5d Mon Sep 17 00:00:00 2001 From: Frankheartusf <104822497+Frankheartusf@users.noreply.github.com> Date: Tue, 3 May 2022 17:31:37 +0800 Subject: [PATCH 056/105] Add files via upload --- 0045.跳跃游戏II.md | 287 +++++++++++++++++++++++++++++++++++++++++ 1 file changed, 287 insertions(+) create mode 100644 0045.跳跃游戏II.md diff --git a/0045.跳跃游戏II.md b/0045.跳跃游戏II.md new file mode 100644 index 00000000..c0d3c3e5 --- /dev/null +++ b/0045.跳跃游戏II.md @@ -0,0 +1,287 @@ +

+ + + +

参与本项目，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们收益！

+ + +> 相对于[贪心算法：跳跃游戏](https://mp.weixin.qq.com/s/606_N9j8ACKCODoCbV1lSA)难了不少，做好心里准备！ + +# 45.跳跃游戏II + +[力扣题目链接](https://leetcode-cn.com/problems/jump-game-ii/) + +给定一个非负整数数组，你最初位于数组的第一个位置。 + +数组中的每个元素代表你在该位置可以跳跃的最大长度。 + +你的目标是使用最少的跳跃次数到达数组的最后一个位置。 + +示例: +* 输入: [2,3,1,1,4] +* 输出: 2 +* 解释: 跳到最后一个位置的最小跳跃数是 2。从下标为 0 跳到下标为 1 的位置，跳 1 步，然后跳 3 步到达数组的最后一个位置。 + +说明: +假设你总是可以到达数组的最后一个位置。 + + +## 思路 + +本题相对于[55.跳跃游戏](https://programmercarl.com/0055.跳跃游戏.html)还是难了不少。 + +但思路是相似的，还是要看最大覆盖范围。 + +本题要计算最小步数，那么就要想清楚什么时候步数才一定要加一呢？ + +贪心的思路，局部最优：当前可移动距离尽可能多走，如果还没到终点，步数再加一。整体最优：一步尽可能多走，从而达到最小步数。 + +思路虽然是这样，但在写代码的时候还不能真的就能跳多远跳远，那样就不知道下一步最远能跳到哪里了。 + +**所以真正解题的时候，要从覆盖范围出发，不管怎么跳，覆盖范围内一定是可以跳到的，以最小的步数增加覆盖范围，覆盖范围一旦覆盖了终点，得到的就是最小步数！** + +**这里需要统计两个覆盖范围，当前这一步的最大覆盖和下一步最大覆盖**。 + +如果移动下标达到了当前这一步的最大覆盖最远距离了，还没有到终点的话，那么就必须再走一步来增加覆盖范围，直到覆盖范围覆盖了终点。 + +如图： + +![45.跳跃游戏II](https://img-blog.csdnimg.cn/20201201232309103.png) + +**图中覆盖范围的意义在于，只要红色的区域，最多两步一定可以到！（不用管具体怎么跳，反正一定可以跳到）** + +## 方法一 + +从图中可以看出来，就是移动下标达到了当前覆盖的最远距离下标时，步数就要加一，来增加覆盖距离。最后的步数就是最少步数。 + +这里还是有个特殊情况需要考虑，当移动下标达到了当前覆盖的最远距离下标时 + +* 如果当前覆盖最远距离下标不是是集合终点，步数就加一，还需要继续走。 +* 如果当前覆盖最远距离下标就是是集合终点，步数不用加一，因为不能再往后走了。 + +C++代码如下：（详细注释） + +```CPP +// 版本一 +class Solution { +public: + int jump(vector& nums) { + if (nums.size() == 1) return 0; + int curDistance = 0; // 当前覆盖最远距离下标 + int ans = 0; // 记录走的最大步数 + int nextDistance = 0; // 下一步覆盖最远距离下标 + for (int i = 0; i < nums.size(); i++) { + nextDistance = max(nums[i] + i, nextDistance); // 更新下一步覆盖最远距离下标 + if (i == curDistance) { // 遇到当前覆盖最远距离下标 + if (curDistance != nums.size() - 1) { // 如果当前覆盖最远距离下标不是终点 + ans++; // 需要走下一步 + curDistance = nextDistance; // 更新当前覆盖最远距离下标（相当于加油了） + if (nextDistance >= nums.size() - 1) break; // 下一步的覆盖范围已经可以达到终点，结束循环 + } else break; // 当前覆盖最远距离下标是集合终点，不用做ans++操作了，直接结束 + } + } + return ans; + } +}; +``` + +## 方法二 + +依然是贪心，思路和方法一差不多，代码可以简洁一些。 + +**针对于方法一的特殊情况，可以统一处理**，即：移动下标只要遇到当前覆盖最远距离的下标，直接步数加一，不考虑是不是终点的情况。 + +想要达到这样的效果，只要让移动下标，最大只能移动到nums.size - 2的地方就可以了。 + +因为当移动下标指向nums.size - 2时： + +* 如果移动下标等于当前覆盖最大距离下标， 需要再走一步（即ans++），因为最后一步一定是可以到的终点。（题目假设总是可以到达数组的最后一个位置），如图： +![45.跳跃游戏II2](https://img-blog.csdnimg.cn/20201201232445286.png) + +* 如果移动下标不等于当前覆盖最大距离下标，说明当前覆盖最远距离就可以直接达到终点了，不需要再走一步。如图： + +![45.跳跃游戏II1](https://img-blog.csdnimg.cn/20201201232338693.png) + +代码如下： + +```CPP +// 版本二 +class Solution { +public: + int jump(vector& nums) { + int curDistance = 0; // 当前覆盖的最远距离下标 + int ans = 0; // 记录走的最大步数 + int nextDistance = 0; // 下一步覆盖的最远距离下标 + for (int i = 0; i < nums.size() - 1; i++) { // 注意这里是小于nums.size() - 1，这是关键所在 + nextDistance = max(nums[i] + i, nextDistance); // 更新下一步覆盖的最远距离下标 + if (i == curDistance) { // 遇到当前覆盖的最远距离下标 + curDistance = nextDistance; // 更新当前覆盖的最远距离下标 + ans++; + } + } + return ans; + } +}; +``` + +可以看出版本二的代码相对于版本一简化了不少！ + +其精髓在于控制移动下标i只移动到nums.size() - 2的位置，所以移动下标只要遇到当前覆盖最远距离的下标，直接步数加一，不用考虑别的了。 + +## 总结 + +相信大家可以发现，这道题目相当于[55.跳跃游戏](https://programmercarl.com/0055.跳跃游戏.html)难了不止一点。 + +但代码又十分简单，贪心就是这么巧妙。 + +理解本题的关键在于：**以最小的步数增加最大的覆盖范围，直到覆盖范围覆盖了终点**，这个范围内最小步数一定可以跳到，不用管具体是怎么跳的，不纠结于一步究竟跳一个单位还是两个单位。 + + +## 其他语言版本 + + +### Java +```Java +// 版本一 +class Solution { + public int jump(int[] nums) { + if (nums == null || nums.length == 0 || nums.length == 1) { + return 0; + } + //记录跳跃的次数 + int count=0; + //当前的覆盖最大区域 + int curDistance = 0; + //最大的覆盖区域 + int maxDistance = 0; + for (int i = 0; i < nums.length; i++) { + //在可覆盖区域内更新最大的覆盖区域 + maxDistance = Math.max(maxDistance,i+nums[i]); + //说明当前一步，再跳一步就到达了末尾 + if (maxDistance>=nums.length-1){ + count++; + break; + } + //走到当前覆盖的最大区域时，更新下一步可达的最大区域 + if (i==curDistance){ + curDistance = maxDistance; + count++; + } + } + return count; + } +} +``` + +```java +// 版本二 +class Solution { + public int jump(int[] nums) { + int result = 0; + // 当前覆盖的最远距离下标 + int end = 0; + // 下一步覆盖的最远距离下标 + int temp = 0; + for (int i = 0; i <= end && end < nums.length - 1; ++i) { + temp = Math.max(temp, i + nums[i]); + // 可达位置的改变次数就是跳跃次数 + if (i == end) { + end = temp; + result++; + } + } + return result; + } +} +``` + +### Python + +```python +class Solution: + def jump(self, nums: List[int]) -> int: + if len(nums) == 1: return 0 + ans = 0 + curDistance = 0 + nextDistance = 0 + for i in range(len(nums)): + nextDistance = max(i + nums[i], nextDistance) + if i == curDistance: + if curDistance != len(nums) - 1: + ans += 1 + curDistance = nextDistance + if nextDistance >= len(nums) - 1: break + return ans +``` + +### Go +```Go +func jump(nums []int) int { + dp := make([]int, len(nums)) + dp[0] = 0//初始第一格跳跃数一定为0 + + for i := 1; i < len(nums); i++ { + dp[i] = i + for j := 0; j < i; j++ { + if nums[j] + j >= i {//nums[j]为起点,j为往右跳的覆盖范围,这行表示从j能跳到i + dp[i] = min(dp[j] + 1, dp[i])//更新最小能到i的跳跃次数 + } + } + } + return dp[len(nums)-1] +} + +func min(a, b int) int { + if a < b { + return a + } else { + return b + } +} +``` + +### Javascript +```Javascript +var jump = function(nums) { + let curIndex = 0 + let nextIndex = 0 + let steps = 0 + for(let i = 0; i < nums.length - 1; i++) { + nextIndex = Math.max(nums[i] + i, nextIndex) + if(i === curIndex) { + curIndex = nextIndex + steps++ + } + } + + return steps +}; +``` + +### TypeScript + +```typescript +function jump(nums: number[]): number { + const length: number = nums.length; + let curFarthestIndex: number = 0, + nextFarthestIndex: number = 0; + let curIndex: number = 0; + let stepNum: number = 0; + while (curIndex < length - 1) { + nextFarthestIndex = Math.max(nextFarthestIndex, curIndex + nums[curIndex]); + if (curIndex === curFarthestIndex) { + curFarthestIndex = nextFarthestIndex; + stepNum++; + } + curIndex++; + } + return stepNum; +}; +``` + + + + + +----------------------- +
From 64d477e3f05c2b7ebb8aa3b5b8ed27578b5a1b31 Mon Sep 17 00:00:00 2001 From: Frankheartusf <2332517004@qq.com> Date: Tue, 3 May 2022 17:50:07 +0800 Subject: [PATCH 057/105] =?UTF-8?q?=E4=BF=AE=E6=94=B9=200045=20=E8=B7=B3?= =?UTF-8?q?=E8=B7=83=E9=97=AE=E9=A2=98II=20golang=E4=BB=A3=E7=A0=81?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 0045.跳跃游戏II.md | 287 -------------------------------- problems/0045.跳跃游戏II.md | 30 ++-- 2 files changed, 19 insertions(+), 298 deletions(-) delete mode 100644 0045.跳跃游戏II.md diff --git a/0045.跳跃游戏II.md b/0045.跳跃游戏II.md deleted file mode 100644 index c0d3c3e5..00000000 --- a/0045.跳跃游戏II.md +++ /dev/null @@ -1,287 +0,0 @@ -

- - - -

参与本项目，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们收益！

- - -> 相对于[贪心算法：跳跃游戏](https://mp.weixin.qq.com/s/606_N9j8ACKCODoCbV1lSA)难了不少，做好心里准备！ - -# 45.跳跃游戏II - -[力扣题目链接](https://leetcode-cn.com/problems/jump-game-ii/) - -给定一个非负整数数组，你最初位于数组的第一个位置。 - -数组中的每个元素代表你在该位置可以跳跃的最大长度。 - -你的目标是使用最少的跳跃次数到达数组的最后一个位置。 - -示例: -* 输入: [2,3,1,1,4] -* 输出: 2 -* 解释: 跳到最后一个位置的最小跳跃数是 2。从下标为 0 跳到下标为 1 的位置，跳 1 步，然后跳 3 步到达数组的最后一个位置。 - -说明: -假设你总是可以到达数组的最后一个位置。 - - -## 思路 - -本题相对于[55.跳跃游戏](https://programmercarl.com/0055.跳跃游戏.html)还是难了不少。 - -但思路是相似的，还是要看最大覆盖范围。 - -本题要计算最小步数，那么就要想清楚什么时候步数才一定要加一呢？ - -贪心的思路，局部最优：当前可移动距离尽可能多走，如果还没到终点，步数再加一。整体最优：一步尽可能多走，从而达到最小步数。 - -思路虽然是这样，但在写代码的时候还不能真的就能跳多远跳远，那样就不知道下一步最远能跳到哪里了。 - -**所以真正解题的时候，要从覆盖范围出发，不管怎么跳，覆盖范围内一定是可以跳到的，以最小的步数增加覆盖范围，覆盖范围一旦覆盖了终点，得到的就是最小步数！** - -**这里需要统计两个覆盖范围，当前这一步的最大覆盖和下一步最大覆盖**。 - -如果移动下标达到了当前这一步的最大覆盖最远距离了，还没有到终点的话，那么就必须再走一步来增加覆盖范围，直到覆盖范围覆盖了终点。 - -如图： - -![45.跳跃游戏II](https://img-blog.csdnimg.cn/20201201232309103.png) - -**图中覆盖范围的意义在于，只要红色的区域，最多两步一定可以到！（不用管具体怎么跳，反正一定可以跳到）** - -## 方法一 - -从图中可以看出来，就是移动下标达到了当前覆盖的最远距离下标时，步数就要加一，来增加覆盖距离。最后的步数就是最少步数。 - -这里还是有个特殊情况需要考虑，当移动下标达到了当前覆盖的最远距离下标时 - -* 如果当前覆盖最远距离下标不是是集合终点，步数就加一，还需要继续走。 -* 如果当前覆盖最远距离下标就是是集合终点，步数不用加一，因为不能再往后走了。 - -C++代码如下：（详细注释） - -```CPP -// 版本一 -class Solution { -public: - int jump(vector& nums) { - if (nums.size() == 1) return 0; - int curDistance = 0; // 当前覆盖最远距离下标 - int ans = 0; // 记录走的最大步数 - int nextDistance = 0; // 下一步覆盖最远距离下标 - for (int i = 0; i < nums.size(); i++) { - nextDistance = max(nums[i] + i, nextDistance); // 更新下一步覆盖最远距离下标 - if (i == curDistance) { // 遇到当前覆盖最远距离下标 - if (curDistance != nums.size() - 1) { // 如果当前覆盖最远距离下标不是终点 - ans++; // 需要走下一步 - curDistance = nextDistance; // 更新当前覆盖最远距离下标（相当于加油了） - if (nextDistance >= nums.size() - 1) break; // 下一步的覆盖范围已经可以达到终点，结束循环 - } else break; // 当前覆盖最远距离下标是集合终点，不用做ans++操作了，直接结束 - } - } - return ans; - } -}; -``` - -## 方法二 - -依然是贪心，思路和方法一差不多，代码可以简洁一些。 - -**针对于方法一的特殊情况，可以统一处理**，即：移动下标只要遇到当前覆盖最远距离的下标，直接步数加一，不考虑是不是终点的情况。 - -想要达到这样的效果，只要让移动下标，最大只能移动到nums.size - 2的地方就可以了。 - -因为当移动下标指向nums.size - 2时： - -* 如果移动下标等于当前覆盖最大距离下标， 需要再走一步（即ans++），因为最后一步一定是可以到的终点。（题目假设总是可以到达数组的最后一个位置），如图： -![45.跳跃游戏II2](https://img-blog.csdnimg.cn/20201201232445286.png) - -* 如果移动下标不等于当前覆盖最大距离下标，说明当前覆盖最远距离就可以直接达到终点了，不需要再走一步。如图： - -![45.跳跃游戏II1](https://img-blog.csdnimg.cn/20201201232338693.png) - -代码如下： - -```CPP -// 版本二 -class Solution { -public: - int jump(vector& nums) { - int curDistance = 0; // 当前覆盖的最远距离下标 - int ans = 0; // 记录走的最大步数 - int nextDistance = 0; // 下一步覆盖的最远距离下标 - for (int i = 0; i < nums.size() - 1; i++) { // 注意这里是小于nums.size() - 1，这是关键所在 - nextDistance = max(nums[i] + i, nextDistance); // 更新下一步覆盖的最远距离下标 - if (i == curDistance) { // 遇到当前覆盖的最远距离下标 - curDistance = nextDistance; // 更新当前覆盖的最远距离下标 - ans++; - } - } - return ans; - } -}; -``` - -可以看出版本二的代码相对于版本一简化了不少！ - -其精髓在于控制移动下标i只移动到nums.size() - 2的位置，所以移动下标只要遇到当前覆盖最远距离的下标，直接步数加一，不用考虑别的了。 - -## 总结 - -相信大家可以发现，这道题目相当于[55.跳跃游戏](https://programmercarl.com/0055.跳跃游戏.html)难了不止一点。 - -但代码又十分简单，贪心就是这么巧妙。 - -理解本题的关键在于：**以最小的步数增加最大的覆盖范围，直到覆盖范围覆盖了终点**，这个范围内最小步数一定可以跳到，不用管具体是怎么跳的，不纠结于一步究竟跳一个单位还是两个单位。 - - -## 其他语言版本 - - -### Java -```Java -// 版本一 -class Solution { - public int jump(int[] nums) { - if (nums == null || nums.length == 0 || nums.length == 1) { - return 0; - } - //记录跳跃的次数 - int count=0; - //当前的覆盖最大区域 - int curDistance = 0; - //最大的覆盖区域 - int maxDistance = 0; - for (int i = 0; i < nums.length; i++) { - //在可覆盖区域内更新最大的覆盖区域 - maxDistance = Math.max(maxDistance,i+nums[i]); - //说明当前一步，再跳一步就到达了末尾 - if (maxDistance>=nums.length-1){ - count++; - break; - } - //走到当前覆盖的最大区域时，更新下一步可达的最大区域 - if (i==curDistance){ - curDistance = maxDistance; - count++; - } - } - return count; - } -} -``` - -```java -// 版本二 -class Solution { - public int jump(int[] nums) { - int result = 0; - // 当前覆盖的最远距离下标 - int end = 0; - // 下一步覆盖的最远距离下标 - int temp = 0; - for (int i = 0; i <= end && end < nums.length - 1; ++i) { - temp = Math.max(temp, i + nums[i]); - // 可达位置的改变次数就是跳跃次数 - if (i == end) { - end = temp; - result++; - } - } - return result; - } -} -``` - -### Python - -```python -class Solution: - def jump(self, nums: List[int]) -> int: - if len(nums) == 1: return 0 - ans = 0 - curDistance = 0 - nextDistance = 0 - for i in range(len(nums)): - nextDistance = max(i + nums[i], nextDistance) - if i == curDistance: - if curDistance != len(nums) - 1: - ans += 1 - curDistance = nextDistance - if nextDistance >= len(nums) - 1: break - return ans -``` - -### Go -```Go -func jump(nums []int) int { - dp := make([]int, len(nums)) - dp[0] = 0//初始第一格跳跃数一定为0 - - for i := 1; i < len(nums); i++ { - dp[i] = i - for j := 0; j < i; j++ { - if nums[j] + j >= i {//nums[j]为起点,j为往右跳的覆盖范围,这行表示从j能跳到i - dp[i] = min(dp[j] + 1, dp[i])//更新最小能到i的跳跃次数 - } - } - } - return dp[len(nums)-1] -} - -func min(a, b int) int { - if a < b { - return a - } else { - return b - } -} -``` - -### Javascript -```Javascript -var jump = function(nums) { - let curIndex = 0 - let nextIndex = 0 - let steps = 0 - for(let i = 0; i < nums.length - 1; i++) { - nextIndex = Math.max(nums[i] + i, nextIndex) - if(i === curIndex) { - curIndex = nextIndex - steps++ - } - } - - return steps -}; -``` - -### TypeScript - -```typescript -function jump(nums: number[]): number { - const length: number = nums.length; - let curFarthestIndex: number = 0, - nextFarthestIndex: number = 0; - let curIndex: number = 0; - let stepNum: number = 0; - while (curIndex < length - 1) { - nextFarthestIndex = Math.max(nextFarthestIndex, curIndex + nums[curIndex]); - if (curIndex === curFarthestIndex) { - curFarthestIndex = nextFarthestIndex; - stepNum++; - } - curIndex++; - } - return stepNum; -}; -``` - - - - - ------------------------ -
diff --git a/problems/0045.跳跃游戏II.md b/problems/0045.跳跃游戏II.md index 4caff042..4e3ab24a 100644 --- a/problems/0045.跳跃游戏II.md +++ b/problems/0045.跳跃游戏II.md @@ -217,18 +217,26 @@ class Solution: ### Go ```Go func jump(nums []int) int { - dp:=make([]int ,len(nums)) - dp[0]=0 + dp := make([]int, len(nums)) + dp[0] = 0//初始第一格跳跃数一定为0 - for i:=1;ii{ - dp[i]=min(dp[j]+1,dp[i]) - } - } - } - return dp[len(nums)-1] + for i := 1; i < len(nums); i++ { + dp[i] = i + for j := 0; j < i; j++ { + if nums[j] + j >= i {//nums[j]为起点,j为往右跳的覆盖范围,这行表示从j能跳到i + dp[i] = min(dp[j] + 1, dp[i])//更新最小能到i的跳跃次数 + } + } + } + return dp[len(nums)-1] +} + +func min(a, b int) int { + if a < b { + return a + } else { + return b + } } ``` From d692ffe034aa1d7af9a4e5d76999ee6a506c23b9 Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Tue, 3 May 2022 21:51:21 +0800 Subject: [PATCH 058/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880416.?= =?UTF-8?q?=E5=88=86=E5=89=B2=E7=AD=89=E5=92=8C=E5=AD=90=E9=9B=86.md?= =?UTF-8?q?=EF=BC=89=EF=BC=9A=E5=A2=9E=E5=8A=A0typescript?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0416.分割等和子集.md | 55 +++++++++++++++++++++++++++++ 1 file changed, 55 insertions(+) diff --git a/problems/0416.分割等和子集.md b/problems/0416.分割等和子集.md index b24fb365..20b7782d 100644 --- a/problems/0416.分割等和子集.md +++ b/problems/0416.分割等和子集.md @@ -416,6 +416,61 @@ var canPartition = function(nums) { }; ``` +TypeScript: + +> 一维数组，简洁 + +```typescript +function canPartition(nums: number[]): boolean { + const sum: number = nums.reduce((pre, cur) => pre + cur); + if (sum % 2 === 1) return false; + const bagSize: number = sum / 2; + const goodsNum: number = nums.length; + const dp: number[] = new Array(bagSize + 1).fill(0); + for (let i = 0; i < goodsNum; i++) { + for (let j = bagSize; j >= nums[i]; j--) { + dp[j] = Math.max(dp[j], dp[j - nums[i]] + nums[i]); + } + } + return dp[bagSize] === bagSize; +}; +``` + +> 二维数组，易懂 + +```typescript +function canPartition(nums: number[]): boolean { + /** + weightArr = nums; + valueArr = nums; + bagSize = sum / 2; (sum为nums各元素总和); + 按照0-1背包处理 + */ + const sum: number = nums.reduce((pre, cur) => pre + cur); + if (sum % 2 === 1) return false; + const bagSize: number = sum / 2; + const weightArr: number[] = nums; + const valueArr: number[] = nums; + const goodsNum: number = weightArr.length; + const dp: number[][] = new Array(goodsNum) + .fill(0) + .map(_ => new Array(bagSize + 1).fill(0)); + for (let i = weightArr[0]; i <= bagSize; i++) { + dp[0][i] = valueArr[0]; + } + for (let i = 1; i < goodsNum; i++) { + for (let j = 0; j <= bagSize; j++) { + if (j < weightArr[i]) { + dp[i][j] = dp[i - 1][j]; + } else { + dp[i][j] = Math.max(dp[i - 1][j], dp[i - 1][j - weightArr[i]] + valueArr[i]); + } + } + } + return dp[goodsNum - 1][bagSize] === bagSize; +}; +``` + From 65cde78559bf18cced2e91b57fba8f92a66437a4 Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Wed, 4 May 2022 12:57:05 +0800 Subject: [PATCH 059/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=881049.?= =?UTF-8?q?=E6=9C=80=E5=90=8E=E4=B8=80=E5=9D=97=E7=9F=B3=E5=A4=B4=E7=9A=84?= =?UTF-8?q?=E9=87=8D=E9=87=8FII.md=EF=BC=89=EF=BC=9A=E5=A2=9E=E5=8A=A0type?= =?UTF-8?q?script=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../1049.最后一块石头的重量II.md | 21 +++++++++++++++++++ 1 file changed, 21 insertions(+) diff --git a/problems/1049.最后一块石头的重量II.md b/problems/1049.最后一块石头的重量II.md index ee0ddef2..3d256c3d 100644 --- a/problems/1049.最后一块石头的重量II.md +++ b/problems/1049.最后一块石头的重量II.md @@ -277,5 +277,26 @@ var lastStoneWeightII = function (stones) { }; ``` +TypeScript： + +```typescript +function lastStoneWeightII(stones: number[]): number { + const sum: number = stones.reduce((pre, cur) => pre + cur); + const bagSize: number = Math.floor(sum / 2); + const weightArr: number[] = stones; + const valueArr: number[] = stones; + const goodsNum: number = weightArr.length; + const dp: number[] = new Array(bagSize + 1).fill(0); + for (let i = 0; i < goodsNum; i++) { + for (let j = bagSize; j >= weightArr[i]; j--) { + dp[j] = Math.max(dp[j], dp[j - weightArr[i]] + valueArr[i]); + } + } + return sum - dp[bagSize] * 2; +}; +``` + + + -----------------------
From b9b7c1530abf7a4b6b68b0f732b5a69cbaae772c Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Wed, 4 May 2022 23:48:12 +0800 Subject: [PATCH 060/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880494.?= =?UTF-8?q?=E7=9B=AE=E6=A0=87=E5=92=8C.md=EF=BC=89=EF=BC=9A=E5=A2=9E?= =?UTF-8?q?=E5=8A=A0typescript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0494.目标和.md | 19 +++++++++++++++++++ 1 file changed, 19 insertions(+) diff --git a/problems/0494.目标和.md b/problems/0494.目标和.md index 99b76834..8ce1f6f1 100644 --- a/problems/0494.目标和.md +++ b/problems/0494.目标和.md @@ -351,6 +351,25 @@ const findTargetSumWays = (nums, target) => { }; ``` +TypeScript： + +```typescript +function findTargetSumWays(nums: number[], target: number): number { + const sum: number = nums.reduce((pre, cur) => pre + cur); + if (Math.abs(target) > sum) return 0; + if ((target + sum) % 2 === 1) return 0; + const bagSize: number = (target + sum) / 2; + const dp: number[] = new Array(bagSize + 1).fill(0); + dp[0] = 1; + for (let i = 0; i < nums.length; i++) { + for (let j = bagSize; j >= nums[i]; j--) { + dp[j] += dp[j - nums[i]]; + } + } + return dp[bagSize]; +}; +``` + ----------------------- From 3c88974f92926f9280d71ea964bdb36b6f097044 Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Thu, 5 May 2022 14:15:00 +0800 Subject: [PATCH 061/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880474.?= =?UTF-8?q?=E4=B8=80=E5=92=8C=E9=9B=B6.md=EF=BC=89=EF=BC=9A=E5=A2=9E?= =?UTF-8?q?=E5=8A=A0typescript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0474.一和零.md | 123 +++++++++++++++++++++++++++++++++++++ 1 file changed, 123 insertions(+) diff --git a/problems/0474.一和零.md b/problems/0474.一和零.md index 964df4a8..d38ce03f 100644 --- a/problems/0474.一和零.md +++ b/problems/0474.一和零.md @@ -323,6 +323,129 @@ const findMaxForm = (strs, m, n) => { }; ``` +TypeScript： + +> 滚动数组，二维数组法 + +```typescript +type BinaryInfo = { numOfZero: number, numOfOne: number }; +function findMaxForm(strs: string[], m: number, n: number): number { + const goodsNum: number = strs.length; + const dp: number[][] = new Array(m + 1).fill(0) + .map(_ => new Array(n + 1).fill(0)); + for (let i = 0; i < goodsNum; i++) { + const { numOfZero, numOfOne } = countBinary(strs[i]); + for (let j = m; j >= numOfZero; j--) { + for (let k = n; k >= numOfOne; k--) { + dp[j][k] = Math.max(dp[j][k], dp[j - numOfZero][k - numOfOne] + 1); + } + } + } + return dp[m][n]; +}; +function countBinary(str: string): BinaryInfo { + let numOfZero: number = 0, + numOfOne: number = 0; + for (let s of str) { + if (s === '0') { + numOfZero++; + } else { + numOfOne++; + } + } + return { numOfZero, numOfOne }; +} +``` + +> 传统背包，三维数组法 + +```typescript +type BinaryInfo = { numOfZero: number, numOfOne: number }; +function findMaxForm(strs: string[], m: number, n: number): number { + /** + dp[i][j][k]: 前i个物品中, 背包的0容量为j, 1容量为k, 最多能放的物品数量 + */ + const goodsNum: number = strs.length; + const dp: number[][][] = new Array(goodsNum).fill(0) + .map(_ => new Array(m + 1) + .fill(0) + .map(_ => new Array(n + 1).fill(0)) + ); + const { numOfZero, numOfOne } = countBinary(strs[0]); + for (let i = numOfZero; i <= m; i++) { + for (let j = numOfOne; j <= n; j++) { + dp[0][i][j] = 1; + } + } + for (let i = 1; i < goodsNum; i++) { + const { numOfZero, numOfOne } = countBinary(strs[i]); + for (let j = 0; j <= m; j++) { + for (let k = 0; k <= n; k++) { + if (j < numOfZero || k < numOfOne) { + dp[i][j][k] = dp[i - 1][j][k]; + } else { + dp[i][j][k] = Math.max(dp[i - 1][j][k], dp[i - 1][j - numOfZero][k - numOfOne] + 1); + } + } + } + } + return dp[dp.length - 1][m][n]; +}; +function countBinary(str: string): BinaryInfo { + let numOfZero: number = 0, + numOfOne: number = 0; + for (let s of str) { + if (s === '0') { + numOfZero++; + } else { + numOfOne++; + } + } + return { numOfZero, numOfOne }; +} +``` + +> 回溯法（会超时） + +```typescript +function findMaxForm(strs: string[], m: number, n: number): number { + /** + 思路：暴力枚举strs的所有子集，记录符合条件子集的最大长度 + */ + let resMax: number = 0; + backTrack(strs, m, n, 0, []); + return resMax; + function backTrack( + strs: string[], m: number, n: number, + startIndex: number, route: string[] + ): void { + if (startIndex === strs.length) return; + for (let i = startIndex, length = strs.length; i < length; i++) { + route.push(strs[i]); + if (isValidSubSet(route, m, n)) { + resMax = Math.max(resMax, route.length); + backTrack(strs, m, n, i + 1, route); + } + route.pop(); + } + } +}; +function isValidSubSet(strs: string[], m: number, n: number): boolean { + let zeroNum: number = 0, + oneNum: number = 0; + strs.forEach(str => { + for (let s of str) { + if (s === '0') { + zeroNum++; + } else { + oneNum++; + } + } + }); + return zeroNum <= m && oneNum <= n; +} +``` + ----------------------- From 1e9cecb665acd511660a9c8c8af48868e20bb871 Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Thu, 5 May 2022 16:53:38 +0800 Subject: [PATCH 062/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=88=E8=83=8C?= =?UTF-8?q?=E5=8C=85=E9=97=AE=E9=A2=98=E7=90=86=E8=AE=BA=E5=9F=BA=E7=A1=80?= =?UTF-8?q?=E5=AE=8C=E5=85=A8=E8=83=8C=E5=8C=85.md=EF=BC=89=EF=BC=9A?= =?UTF-8?q?=E5=A2=9E=E5=8A=A0typescript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../背包问题理论基础完全背包.md | 21 +++++++++++++++++++ 1 file changed, 21 insertions(+) diff --git a/problems/背包问题理论基础完全背包.md b/problems/背包问题理论基础完全背包.md index faa1dc46..936a80c7 100644 --- a/problems/背包问题理论基础完全背包.md +++ b/problems/背包问题理论基础完全背包.md @@ -340,6 +340,27 @@ function test_completePack2() { } ``` +TypeScript： + +```typescript +// 先遍历物品，再遍历背包容量 +function test_CompletePack(): void { + const weight: number[] = [1, 3, 4]; + const value: number[] = [15, 20, 30]; + const bagSize: number = 4; + const dp: number[] = new Array(bagSize + 1).fill(0); + for (let i = 0; i < weight.length; i++) { + for (let j = weight[i]; j <= bagSize; j++) { + dp[j] = Math.max(dp[j], dp[j - weight[i]] + value[i]); + } + } + console.log(dp); +} +test_CompletePack(); +``` + + + -----------------------
From b82984e24b23940aa95da2a0283ae4cf3c706f7a Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Thu, 5 May 2022 18:11:41 +0800 Subject: [PATCH 063/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880518.?= =?UTF-8?q?=E9=9B=B6=E9=92=B1=E5=85=91=E6=8D=A2II.md=EF=BC=89=EF=BC=9A?= =?UTF-8?q?=E5=A2=9E=E5=8A=A0typescript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0518.零钱兑换II.md | 15 +++++++++++++++ 1 file changed, 15 insertions(+) diff --git a/problems/0518.零钱兑换II.md b/problems/0518.零钱兑换II.md index e72c5f85..8faf6698 100644 --- a/problems/0518.零钱兑换II.md +++ b/problems/0518.零钱兑换II.md @@ -258,6 +258,21 @@ const change = (amount, coins) => { } ``` +TypeScript： + +```typescript +function change(amount: number, coins: number[]): number { + const dp: number[] = new Array(amount + 1).fill(0); + dp[0] = 1; + for (let i = 0, length = coins.length; i < length; i++) { + for (let j = coins[i]; j <= amount; j++) { + dp[j] += dp[j - coins[i]]; + } + } + return dp[amount]; +}; +``` + ----------------------- From cb7bf67a4e3a277ecf53ce37d0556c8e2e65f9f3 Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Fri, 6 May 2022 11:08:47 +0800 Subject: [PATCH 064/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880377.?= =?UTF-8?q?=E7=BB=84=E5=90=88=E6=80=BB=E5=92=8C=E2=85=A3.md=EF=BC=89?= =?UTF-8?q?=EF=BC=9A=E5=A2=9E=E5=8A=A0typescript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0377.组合总和Ⅳ.md | 20 ++++++++++++++++++++ 1 file changed, 20 insertions(+) diff --git a/problems/0377.组合总和Ⅳ.md b/problems/0377.组合总和Ⅳ.md index aaf27e61..1d808a3a 100644 --- a/problems/0377.组合总和Ⅳ.md +++ b/problems/0377.组合总和Ⅳ.md @@ -221,7 +221,27 @@ const combinationSum4 = (nums, target) => { }; ``` +TypeScript： + +```typescript +function combinationSum4(nums: number[], target: number): number { + const dp: number[] = new Array(target + 1).fill(0); + dp[0] = 1; + // 遍历背包 + for (let i = 1; i <= target; i++) { + // 遍历物品 + for (let j = 0, length = nums.length; j < length; j++) { + if (i >= nums[j]) { + dp[i] += dp[i - nums[j]]; + } + } + } + return dp[target]; +}; +``` + Rust + ```Rust impl Solution { pub fn combination_sum4(nums: Vec, target: i32) -> i32 { From 100a4e2b46abf913dacc62637da7faf27bfe060e Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Fri, 6 May 2022 11:56:19 +0800 Subject: [PATCH 065/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880070.?= =?UTF-8?q?=E7=88=AC=E6=A5=BC=E6=A2=AF=E5=AE=8C=E5=85=A8=E8=83=8C=E5=8C=85?= =?UTF-8?q?=E7=89=88=E6=9C=AC.md=EF=BC=89=EF=BC=9A=E5=A2=9E=E5=8A=A0typesc?= =?UTF-8?q?ript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0070.爬楼梯完全背包版本.md | 22 ++++++++++++++++++++ 1 file changed, 22 insertions(+) diff --git a/problems/0070.爬楼梯完全背包版本.md b/problems/0070.爬楼梯完全背包版本.md index 2286de2d..0f482bb7 100644 --- a/problems/0070.爬楼梯完全背包版本.md +++ b/problems/0070.爬楼梯完全背包版本.md @@ -199,6 +199,28 @@ var climbStairs = function(n) { }; ``` +TypeScript： + +```typescript +function climbStairs(n: number): number { + const m: number = 2; // 本题m为2 + const dp: number[] = new Array(n + 1).fill(0); + dp[0] = 1; + // 遍历背包 + for (let i = 1; i <= n; i++) { + // 遍历物品 + for (let j = 1; j <= m; j++) { + if (j <= i) { + dp[i] += dp[i - j]; + } + } + } + return dp[n]; +}; +``` + + + -----------------------
From cb6f015186ae9d6af13cddf9b929f7bbdfe9bc29 Mon Sep 17 00:00:00 2001 From: cy948 <67412196+cy948@users.noreply.github.com> Date: Fri, 6 May 2022 14:43:58 +0800 Subject: [PATCH 066/105] =?UTF-8?q?Update=200112.=E8=B7=AF=E5=BE=84?= =?UTF-8?q?=E6=80=BB=E5=92=8C.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 修复了 0113.路径总和-ii 中Java递归解法中的类名大小写问题。 --- problems/0112.路径总和.md | 12 ++++++------ 1 file changed, 6 insertions(+), 6 deletions(-) diff --git a/problems/0112.路径总和.md b/problems/0112.路径总和.md index 41463ec1..2fdd7741 100644 --- a/problems/0112.路径总和.md +++ b/problems/0112.路径总和.md @@ -377,22 +377,22 @@ class solution { ```java class solution { - public list> pathsum(treenode root, int targetsum) { - list> res = new arraylist<>(); + public List> pathsum(TreeNode root, int targetsum) { + List> res = new ArrayList<>(); if (root == null) return res; // 非空判断 - - list path = new linkedlist<>(); + + List path = new LinkedList<>(); preorderdfs(root, targetsum, res, path); return res; } - public void preorderdfs(treenode root, int targetsum, list> res, list path) { + public void preorderdfs(TreeNode root, int targetsum, List> res, List path) { path.add(root.val); // 遇到了叶子节点 if (root.left == null && root.right == null) { // 找到了和为 targetsum 的路径 if (targetsum - root.val == 0) { - res.add(new arraylist<>(path)); + res.add(new ArrayList<>(path)); } return; // 如果和不为 targetsum，返回 } From 00b5e2aa64f49ce9273aca6aa8074d98fe4e489a Mon Sep 17 00:00:00 2001 From: eat to 160 pounds <2915390277@qq.com> Date: Fri, 6 May 2022 15:06:32 +0800 Subject: [PATCH 067/105] =?UTF-8?q?131=E5=88=86=E5=89=B2=E5=AD=97=E7=AC=A6?= =?UTF-8?q?=E4=B8=B2=E6=9B=B4=E6=B8=85=E6=A5=9A=E7=9A=84typescript?= =?UTF-8?q?=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0131.分割回文串.md | 41 ++++++++++++++++++-------------- 1 file changed, 23 insertions(+), 18 deletions(-) diff --git a/problems/0131.分割回文串.md b/problems/0131.分割回文串.md index 10b747cb..7a702898 100644 --- a/problems/0131.分割回文串.md +++ b/problems/0131.分割回文串.md @@ -454,31 +454,36 @@ var partition = function(s) { ```typescript function partition(s: string): string[][] { - function isPalindromeStr(s: string, left: number, right: number): boolean { - while (left < right) { - if (s[left++] !== s[right--]) { - return false; + const res: string[][] = [] + const path: string[] = [] + const isHuiwen = ( + str: string, + startIndex: number, + endIndex: number + ): boolean => { + for (; startIndex < endIndex; startIndex++, endIndex--) { + if (str[startIndex] !== str[endIndex]) { + return false } } - return true; + return true } - function backTracking(s: string, startIndex: number, route: string[]): void { - let length: number = s.length; - if (length === startIndex) { - resArr.push(route.slice()); - return; + const rec = (str: string, index: number): void => { + if (index >= str.length) { + res.push([...path]) + return } - for (let i = startIndex; i < length; i++) { - if (isPalindromeStr(s, startIndex, i)) { - route.push(s.slice(startIndex, i + 1)); - backTracking(s, i + 1, route); - route.pop(); + for (let i = index; i < str.length; i++) { + if (!isHuiwen(str, index, i)) { + continue } + path.push(str.substring(index, i + 1)) + rec(str, i + 1) + path.pop() } } - const resArr: string[][] = []; - backTracking(s, 0, []); - return resArr; + rec(s, 0) + return res }; ``` From 82a58bc20249bb248dbbb0d90771db0ea60bf017 Mon Sep 17 00:00:00 2001 From: Jamcy123 <1219502823@qq.com> Date: Fri, 6 May 2022 15:49:43 +0800 Subject: [PATCH 068/105] =?UTF-8?q?1005.K=E6=AC=A1=E5=8F=96=E5=8F=8D?= =?UTF-8?q?=E5=90=8E=E6=9C=80=E5=A4=A7=E5=8C=96=E7=9A=84=E6=95=B0=E7=BB=84?= =?UTF-8?q?=E5=92=8C?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../1005.K次取反后最大化的数组和.md | 16 ++++++++++++++++ 1 file changed, 16 insertions(+) diff --git a/problems/1005.K次取反后最大化的数组和.md b/problems/1005.K次取反后最大化的数组和.md index 79767deb..202534da 100644 --- a/problems/1005.K次取反后最大化的数组和.md +++ b/problems/1005.K次取反后最大化的数组和.md @@ -209,6 +209,22 @@ var largestSumAfterKNegations = function(nums, k) { return a + b }) }; + +// 版本二 (优化: 一次遍历) +var largestSumAfterKNegations = function(nums, k) { + nums.sort((a, b) => Math.abs(b) - Math.abs(a)); // 排序 + let sum = 0; + for(let i = 0; i < nums.length; i++) { + if(nums[i] < 0 && k-- > 0) { // 负数取反（k 数量足够时） + nums[i] = -nums[i]; + } + sum += nums[i]; // 求和 + } + if(k % 2 > 0) { // k 有多余的（k若消耗完则应为 -1） + sum -= 2 * nums[nums.length - 1]; // 减去两倍的最小值（因为之前加过一次） + } + return sum; +}; ``` From 08ae851e75f735f45f033a0c5455aef198d3cc92 Mon Sep 17 00:00:00 2001 From: wang <472146630@qq.com> Date: Fri, 6 May 2022 20:45:29 +0800 Subject: [PATCH 069/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A01035.=E4=B8=8D?= =?UTF-8?q?=E7=9B=B8=E4=BA=A4=E7=9A=84=E7=BA=BF=E5=92=8C1143.=E6=9C=80?= =?UTF-8?q?=E9=95=BF=E5=85=AC=E5=85=B1=E5=AD=90=E5=BA=8F=E5=88=97?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/1035.不相交的线.md | 20 ++++++- problems/1143.最长公共子序列.md | 80 ++++++++++++++++---------- 2 files changed, 70 insertions(+), 30 deletions(-) diff --git a/problems/1035.不相交的线.md b/problems/1035.不相交的线.md index 0602e111..279ed816 100644 --- a/problems/1035.不相交的线.md +++ b/problems/1035.不相交的线.md @@ -111,7 +111,6 @@ class Solution: Golang: ```go - func maxUncrossedLines(A []int, B []int) int { m, n := len(A), len(B) dp := make([][]int, m+1) @@ -140,7 +139,26 @@ func max(a, b int) int { } ``` +Rust: +```rust +pub fn max_uncrossed_lines(nums1: Vec, nums2: Vec) -> i32 { + let (n, m) = (nums1.len(), nums2.len()); + let mut last = vec![0; m + 1]; // 记录滚动数组 + let mut dp = vec![0; m + 1]; + for i in 1..=n { + dp.swap_with_slice(&mut last); + for j in 1..=m { + if nums1[i - 1] == nums2[j - 1] { + dp[j] = last[j - 1] + 1; + } else { + dp[j] = last[j].max(dp[j - 1]); + } + } + } + dp[m] +} +``` JavaScript： diff --git a/problems/1143.最长公共子序列.md b/problems/1143.最长公共子序列.md index fdcc7619..ecedf89b 100644 --- a/problems/1143.最长公共子序列.md +++ b/problems/1143.最长公共子序列.md @@ -4,40 +4,40 @@

参与本项目，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们收益！

-## 1143.最长公共子序列 +## 1143.最长公共子序列 [力扣题目链接](https://leetcode-cn.com/problems/longest-common-subsequence/) -给定两个字符串 text1 和 text2，返回这两个字符串的最长公共子序列的长度。 +给定两个字符串 text1 和 text2，返回这两个字符串的最长公共子序列的长度。 -一个字符串的 子序列 是指这样一个新的字符串：它是由原字符串在不改变字符的相对顺序的情况下删除某些字符（也可以不删除任何字符）后组成的新字符串。 +一个字符串的 子序列 是指这样一个新的字符串：它是由原字符串在不改变字符的相对顺序的情况下删除某些字符（也可以不删除任何字符）后组成的新字符串。 -例如，"ace" 是 "abcde" 的子序列，但 "aec" 不是 "abcde" 的子序列。两个字符串的「公共子序列」是这两个字符串所共同拥有的子序列。 +例如，"ace" 是 "abcde" 的子序列，但 "aec" 不是 "abcde" 的子序列。两个字符串的「公共子序列」是这两个字符串所共同拥有的子序列。 -若这两个字符串没有公共子序列，则返回 0。 +若这两个字符串没有公共子序列，则返回 0。 -示例 1: +示例 1: -输入：text1 = "abcde", text2 = "ace" -输出：3 -解释：最长公共子序列是 "ace"，它的长度为 3。 +输入：text1 = "abcde", text2 = "ace" +输出：3 +解释：最长公共子序列是 "ace"，它的长度为 3。 -示例 2: -输入：text1 = "abc", text2 = "abc" -输出：3 -解释：最长公共子序列是 "abc"，它的长度为 3。 +示例 2: +输入：text1 = "abc", text2 = "abc" +输出：3 +解释：最长公共子序列是 "abc"，它的长度为 3。 -示例 3: -输入：text1 = "abc", text2 = "def" -输出：0 -解释：两个字符串没有公共子序列，返回 0。 +示例 3: +输入：text1 = "abc", text2 = "def" +输出：0 +解释：两个字符串没有公共子序列，返回 0。 -提示: +提示: * 1 <= text1.length <= 1000 * 1 <= text2.length <= 1000 输入的字符串只含有小写英文字符。 -## 思路 +## 思路 本题和[动态规划：718. 最长重复子数组](https://programmercarl.com/0718.最长重复子数组.html)区别在于这里不要求是连续的了，但要有相对顺序，即："ace" 是 "abcde" 的子序列，但 "aec" 不是 "abcde" 的子序列。 @@ -45,21 +45,21 @@ 1. 确定dp数组（dp table）以及下标的含义 -dp[i][j]：长度为[0, i - 1]的字符串text1与长度为[0, j - 1]的字符串text2的最长公共子序列为dp[i][j] +dp[i][j]：长度为[0, i - 1]的字符串text1与长度为[0, j - 1]的字符串text2的最长公共子序列为dp[i][j] -有同学会问：为什么要定义长度为[0, i - 1]的字符串text1，定义为长度为[0, i]的字符串text1不香么？ +有同学会问：为什么要定义长度为[0, i - 1]的字符串text1，定义为长度为[0, i]的字符串text1不香么？ 这样定义是为了后面代码实现方便，如果非要定义为为长度为[0, i]的字符串text1也可以，大家可以试一试！ 2. 确定递推公式 -主要就是两大情况： text1[i - 1] 与 text2[j - 1]相同，text1[i - 1] 与 text2[j - 1]不相同 +主要就是两大情况： text1[i - 1] 与 text2[j - 1]相同，text1[i - 1] 与 text2[j - 1]不相同 -如果text1[i - 1] 与 text2[j - 1]相同，那么找到了一个公共元素，所以dp[i][j] = dp[i - 1][j - 1] + 1; +如果text1[i - 1] 与 text2[j - 1]相同，那么找到了一个公共元素，所以dp[i][j] = dp[i - 1][j - 1] + 1; 如果text1[i - 1] 与 text2[j - 1]不相同，那就看看text1[0, i - 2]与text2[0, j - 1]的最长公共子序列 和 text1[0, i - 1]与text2[0, j - 2]的最长公共子序列，取最大的。 -即：dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]); +即：dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]); 代码如下： @@ -71,9 +71,9 @@ if (text1[i - 1] == text2[j - 1]) { } ``` -3. dp数组如何初始化 +3. dp数组如何初始化 -先看看dp[i][0]应该是多少呢？ +先看看dp[i][0]应该是多少呢？ test1[0, i-1]和空串的最长公共子序列自然是0，所以dp[i][0] = 0; @@ -101,7 +101,7 @@ vector> dp(text1.size() + 1, vector(text2.size() + 1, 0)); ![1143.最长公共子序列1](https://img-blog.csdnimg.cn/20210210150215918.jpg) -最后红框dp[text1.size()][text2.size()]为最终结果 +最后红框dp[text1.size()][text2.size()]为最终结果 以上分析完毕，C++代码如下： @@ -158,7 +158,7 @@ class Solution: for i in range(1, len2): for j in range(1, len1): # 开始列出状态转移方程 if text1[j-1] == text2[i-1]: - dp[i][j] = dp[i-1][j-1]+1 + dp[i][j] = dp[i-1][j-1]+1 else: dp[i][j] = max(dp[i-1][j], dp[i][j-1]) return dp[-1][-1] @@ -189,10 +189,32 @@ func longestCommonSubsequence(text1 string, text2 string) int { func max(a,b int)int { if a>b{ - return a + return a } return b } + +``` + +Rust: +```rust +pub fn longest_common_subsequence(text1: String, text2: String) -> i32 { + let (n, m) = (text1.len(), text2.len()); + let (s1, s2) = (text1.as_bytes(), text2.as_bytes()); + let mut dp = vec![0; m + 1]; + let mut last = vec![0; m + 1]; + for i in 1..=n { + dp.swap_with_slice(&mut last); + for j in 1..=m { + dp[j] = if s1[i - 1] == s2[j - 1] { + last[j - 1] + 1 + } else { + last[j].max(dp[j - 1]) + }; + } + } + dp[m] +} ``` Javascript： From 132e57e2ddd801d7360401e746d945957c7e0801 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E5=AD=9F=E4=BB=A4=E4=BB=A4?= Date: Sat, 7 May 2022 09:57:27 +0800 Subject: [PATCH 070/105] =?UTF-8?q?Update=200332.=E9=87=8D=E6=96=B0?= =?UTF-8?q?=E5=AE=89=E6=8E=92=E8=A1=8C=E7=A8=8B.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 恢复0332 原go代码 --- problems/0332.重新安排行程.md | 58 +++++++++++++++++++++++++++++ 1 file changed, 58 insertions(+) diff --git a/problems/0332.重新安排行程.md b/problems/0332.重新安排行程.md index d087d97b..c71b2a93 100644 --- a/problems/0332.重新安排行程.md +++ b/problems/0332.重新安排行程.md @@ -342,6 +342,64 @@ class Solution: return path ``` +### GO +```go +type pair struct { + target string + visited bool +} +type pairs []*pair + +func (p pairs) Len() int { + return len(p) +} +func (p pairs) Swap(i, j int) { + p[i], p[j] = p[j], p[i] +} +func (p pairs) Less(i, j int) bool { + return p[i].target < p[j].target +} + +func findItinerary(tickets [][]string) []string { + result := []string{} + // map[出发机场] pair{目的地,是否被访问过} + targets := make(map[string]pairs) + for _, ticket := range tickets { + if targets[ticket[0]] == nil { + targets[ticket[0]] = make(pairs, 0) + } + targets[ticket[0]] = append(targets[ticket[0]], &pair{target: ticket[1], visited: false}) + } + for k, _ := range targets { + sort.Sort(targets[k]) + } + result = append(result, "JFK") + var backtracking func() bool + backtracking = func() bool { + if len(tickets)+1 == len(result) { + return true + } + // 取出起飞航班对应的目的地 + for _, pair := range targets[result[len(result)-1]] { + if pair.visited == false { + result = append(result, pair.target) + pair.visited = true + if backtracking() { + return true + } + result = result[:len(result)-1] + pair.visited = false + } + } + return false + } + + backtracking() + + return result +} +``` + ### C语言 ```C From 29f41716f971506a001e8eb369b3f62964b43e8e Mon Sep 17 00:00:00 2001 From: GitHubQAQ <31883473+GitHubQAQ@users.noreply.github.com> Date: Sat, 7 May 2022 10:12:58 +0800 Subject: [PATCH 071/105] =?UTF-8?q?Update=200093.=E5=A4=8D=E5=8E=9FIP?= =?UTF-8?q?=E5=9C=B0=E5=9D=80.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 1. 优化C++版本剪枝 当字符串长度小于4时，无法组成正确的IP地址，直接返回。 关于判断是否为数字的部分，是否有必要（题目明确说明s仅由数字组成） [从解题角度可以删去，保留的话增加容错，个人意见] --- problems/0093.复原IP地址.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0093.复原IP地址.md b/problems/0093.复原IP地址.md index 7910fc50..6401824b 100644 --- a/problems/0093.复原IP地址.md +++ b/problems/0093.复原IP地址.md @@ -227,7 +227,7 @@ private: public: vector restoreIpAddresses(string s) { result.clear(); - if (s.size() > 12) return result; // 算是剪枝了 + if (s.size() < 4 || s.size() > 12) return result; // 算是剪枝了 backtracking(s, 0, 0); return result; } From da464016574f14c43fe1e9f7c0d808ed7ea304fb Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Sun, 8 May 2022 10:29:33 +0800 Subject: [PATCH 072/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880322.?= =?UTF-8?q?=E9=9B=B6=E9=92=B1=E5=85=91=E6=8D=A2.md=EF=BC=89=EF=BC=9A?= =?UTF-8?q?=E5=A2=9E=E5=8A=A0typescript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0322.零钱兑换.md | 14 ++++++++++++++ 1 file changed, 14 insertions(+) diff --git a/problems/0322.零钱兑换.md b/problems/0322.零钱兑换.md index 3a8d0662..1df8d613 100644 --- a/problems/0322.零钱兑换.md +++ b/problems/0322.零钱兑换.md @@ -322,7 +322,21 @@ const coinChange = (coins, amount) => { } ``` +TypeScript： +```typescript +function coinChange(coins: number[], amount: number): number { + const dp: number[] = new Array(amount + 1).fill(Infinity); + dp[0] = 0; + for (let i = 0; i < coins.length; i++) { + for (let j = coins[i]; j <= amount; j++) { + if (dp[j - coins[i]] === Infinity) continue; + dp[j] = Math.min(dp[j], dp[j - coins[i]] + 1); + } + } + return dp[amount] === Infinity ? -1 : dp[amount]; +}; +``` -----------------------
From 2722fe7b5eceec7d746f94a63683430205839469 Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Sun, 8 May 2022 11:07:39 +0800 Subject: [PATCH 073/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880279.?= =?UTF-8?q?=E5=AE=8C=E5=85=A8=E5=B9=B3=E6=96=B9=E6=95=B0.md=EF=BC=89?= =?UTF-8?q?=EF=BC=9A=E5=A2=9E=E5=8A=A0typescript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0279.完全平方数.md | 19 +++++++++++++++++++ 1 file changed, 19 insertions(+) diff --git a/problems/0279.完全平方数.md b/problems/0279.完全平方数.md index 9bad2085..5b15639c 100644 --- a/problems/0279.完全平方数.md +++ b/problems/0279.完全平方数.md @@ -355,5 +355,24 @@ var numSquares2 = function(n) { }; ``` +TypeScript： + +```typescript +function numSquares(n: number): number { + const goodsNum: number = Math.floor(Math.sqrt(n)); + const dp: number[] = new Array(n + 1).fill(Infinity); + dp[0] = 0; + for (let i = 1; i <= goodsNum; i++) { + const tempVal: number = i * i; + for (let j = tempVal; j <= n; j++) { + dp[j] = Math.min(dp[j], dp[j - tempVal] + 1); + } + } + return dp[n]; +}; +``` + + + -----------------------
From 9b6a447674f3290541e46088962e1e3aa764d827 Mon Sep 17 00:00:00 2001 From: Hayden-Chang <62008508+Hayden-Chang@users.noreply.github.com> Date: Sun, 8 May 2022 14:20:09 +0800 Subject: [PATCH 074/105] explain the reason for reverse traversal --- problems/背包理论基础01背包-2.md | 1 + 1 file changed, 1 insertion(+) diff --git a/problems/背包理论基础01背包-2.md b/problems/背包理论基础01背包-2.md index dabdfb2d..ea7c53ad 100644 --- a/problems/背包理论基础01背包-2.md +++ b/problems/背包理论基础01背包-2.md @@ -136,6 +136,7 @@ dp[1] = dp[1 - weight[0]] + value[0] = 15 不可以！ 因为一维dp的写法，背包容量一定是要倒序遍历（原因上面已经讲了），如果遍历背包容量放在上一层，那么每个dp[j]就只会放入一个物品，即：背包里只放入了一个物品。 +倒叙遍历的原因是，本质上还是一个对二维数组的遍历，并且右下角的值依赖上一层左上角的值，因此需要保证左边的值仍然是上一层的，从右向左覆盖。 （这里如果读不懂，就在回想一下dp[j]的定义，或者就把两个for循环顺序颠倒一下试试！） From 80167289e4895c4116b61a5c2a4fa22e9e2e88b5 Mon Sep 17 00:00:00 2001 From: Hayden-Chang <62008508+Hayden-Chang@users.noreply.github.com> Date: Sun, 8 May 2022 14:24:09 +0800 Subject: [PATCH 075/105] =?UTF-8?q?Update=20=E8=83=8C=E5=8C=85=E7=90=86?= =?UTF-8?q?=E8=AE=BA=E5=9F=BA=E7=A1=8001=E8=83=8C=E5=8C=85-2.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/背包理论基础01背包-2.md | 3 ++- 1 file changed, 2 insertions(+), 1 deletion(-) diff --git a/problems/背包理论基础01背包-2.md b/problems/背包理论基础01背包-2.md index ea7c53ad..eae01158 100644 --- a/problems/背包理论基础01背包-2.md +++ b/problems/背包理论基础01背包-2.md @@ -136,7 +136,8 @@ dp[1] = dp[1 - weight[0]] + value[0] = 15 不可以！ 因为一维dp的写法，背包容量一定是要倒序遍历（原因上面已经讲了），如果遍历背包容量放在上一层，那么每个dp[j]就只会放入一个物品，即：背包里只放入了一个物品。 -倒叙遍历的原因是，本质上还是一个对二维数组的遍历，并且右下角的值依赖上一层左上角的值，因此需要保证左边的值仍然是上一层的，从右向左覆盖。 + +倒序遍历的原因是，本质上还是一个对二维数组的遍历，并且右下角的值依赖上一层左上角的值，因此需要保证左边的值仍然是上一层的，从右向左覆盖。 （这里如果读不懂，就在回想一下dp[j]的定义，或者就把两个for循环顺序颠倒一下试试！） From 2fb34b30b38ee73624d65d183955cef3a80caced Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Sun, 8 May 2022 18:28:32 +0800 Subject: [PATCH 076/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880139.?= =?UTF-8?q?=E5=8D=95=E8=AF=8D=E6=8B=86=E5=88=86.md=EF=BC=89:=E5=A2=9E?= =?UTF-8?q?=E5=8A=A0typescript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0139.单词拆分.md | 42 +++++++++++++++++++++++++++++++++++ 1 file changed, 42 insertions(+) diff --git a/problems/0139.单词拆分.md b/problems/0139.单词拆分.md index ac834f04..5b4e92b9 100644 --- a/problems/0139.单词拆分.md +++ b/problems/0139.单词拆分.md @@ -345,6 +345,48 @@ const wordBreak = (s, wordDict) => { } ``` +TypeScript： + +> 动态规划 + +```typescript +function wordBreak(s: string, wordDict: string[]): boolean { + const dp: boolean[] = new Array(s.length + 1).fill(false); + dp[0] = true; + for (let i = 1; i <= s.length; i++) { + for (let j = 0; j < i; j++) { + const tempStr: string = s.slice(j, i); + if (wordDict.includes(tempStr) && dp[j] === true) { + dp[i] = true; + break; + } + } + } + return dp[s.length]; +}; +``` + +> 记忆化回溯 + +```typescript +function wordBreak(s: string, wordDict: string[]): boolean { + // 只需要记忆结果为false的情况 + const memory: boolean[] = []; + return backTracking(s, wordDict, 0, memory); + function backTracking(s: string, wordDict: string[], startIndex: number, memory: boolean[]): boolean { + if (startIndex >= s.length) return true; + if (memory[startIndex] === false) return false; + for (let i = startIndex + 1, length = s.length; i <= length; i++) { + const str: string = s.slice(startIndex, i); + if (wordDict.includes(str) && backTracking(s, wordDict, i, memory)) + return true; + } + memory[startIndex] = false; + return false; + } +}; +``` + ----------------------- From 21475ad21864ce33fb5a6a113581f2c346be78b6 Mon Sep 17 00:00:00 2001 From: languagege Date: Mon, 9 May 2022 18:51:39 +0800 Subject: [PATCH 077/105] =?UTF-8?q?=E4=BF=AE=E6=AD=A3=E4=BA=86=E5=85=B6?= =?UTF-8?q?=E4=B8=AD=E4=B8=80=E4=B8=AA=E9=94=99=E5=88=AB=E5=AD=97?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0027.移除元素.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0027.移除元素.md b/problems/0027.移除元素.md index 590cf0b9..3a93ac88 100644 --- a/problems/0027.移除元素.md +++ b/problems/0027.移除元素.md @@ -81,7 +81,7 @@ public: **双指针法（快慢指针法）在数组和链表的操作中是非常常见的，很多考察数组、链表、字符串等操作的面试题，都使用双指针法。** -后序都会一一介绍到，本题代码如下： +后续都会一一介绍到，本题代码如下： ```CPP // 时间复杂度：O(n) From 75198263f91484020c4a4dde421d9ecd989422dd Mon Sep 17 00:00:00 2001 From: changjunkui <506678275@qq.com> Date: Tue, 10 May 2022 08:16:53 +0800 Subject: [PATCH 078/105] =?UTF-8?q?Update=200019.=E5=88=A0=E9=99=A4?= =?UTF-8?q?=E9=93=BE=E8=A1=A8=E7=9A=84=E5=80=92=E6=95=B0=E7=AC=ACN?= =?UTF-8?q?=E4=B8=AA=E8=8A=82=E7=82=B9.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 方面改为方便 --- problems/0019.删除链表的倒数第N个节点.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0019.删除链表的倒数第N个节点.md b/problems/0019.删除链表的倒数第N个节点.md index 813e9b02..b3030a81 100644 --- a/problems/0019.删除链表的倒数第N个节点.md +++ b/problems/0019.删除链表的倒数第N个节点.md @@ -39,7 +39,7 @@ 分为如下几步： -* 首先这里我推荐大家使用虚拟头结点，这样方面处理删除实际头结点的逻辑，如果虚拟头结点不清楚，可以看这篇： [链表：听说用虚拟头节点会方便很多？](https://programmercarl.com/0203.移除链表元素.html) +* 首先这里我推荐大家使用虚拟头结点，这样方便处理删除实际头结点的逻辑，如果虚拟头结点不清楚，可以看这篇： [链表：听说用虚拟头节点会方便很多？](https://programmercarl.com/0203.移除链表元素.html) * 定义fast指针和slow指针，初始值为虚拟头结点，如图： From 8bab33d0bba8a080fb614bec6be3b0a5961d7727 Mon Sep 17 00:00:00 2001 From: wang <472146630@qq.com> Date: Tue, 10 May 2022 21:27:54 +0800 Subject: [PATCH 079/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200300.=E6=9C=80?= =?UTF-8?q?=E9=95=BF=E4=B8=8A=E5=8D=87=E5=AD=90=E5=BA=8F=E5=88=97=E3=80=81?= =?UTF-8?q?0322.=E9=9B=B6=E9=92=B1=E5=85=91=E6=8D=A2=E3=80=810518.?= =?UTF-8?q?=E9=9B=B6=E9=92=B1=E5=85=91=E6=8D=A2II=20=E5=92=8C0674.?= =?UTF-8?q?=E6=9C=80=E9=95=BF=E8=BF=9E=E7=BB=AD=E9=80=92=E5=A2=9E=E5=BA=8F?= =?UTF-8?q?=E5=88=97=E7=9A=84rust=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0300.最长上升子序列.md | 17 +++++++++++++++++ problems/0322.零钱兑换.md | 20 +++++++++++++++++++- problems/0518.零钱兑换II.md | 16 ++++++++++++++++ problems/0674.最长连续递增序列.md | 19 +++++++++++++++++++ 4 files changed, 71 insertions(+), 1 deletion(-) diff --git a/problems/0300.最长上升子序列.md b/problems/0300.最长上升子序列.md index dfdd5125..f53d19a1 100644 --- a/problems/0300.最长上升子序列.md +++ b/problems/0300.最长上升子序列.md @@ -168,6 +168,23 @@ func lengthOfLIS(nums []int ) int { } ``` +Rust: +```rust +pub fn length_of_lis(nums: Vec) -> i32 { + let mut dp = vec![1; nums.len() + 1]; + let mut result = 1; + for i in 1..nums.len() { + for j in 0..i { + if nums[j] < nums[i] { + dp[i] = dp[i].max(dp[j] + 1); + } + result = result.max(dp[i]); + } + } + result +} +``` + Javascript ```javascript const lengthOfLIS = (nums) => { diff --git a/problems/0322.零钱兑换.md b/problems/0322.零钱兑换.md index 3a8d0662..43c735be 100644 --- a/problems/0322.零钱兑换.md +++ b/problems/0322.零钱兑换.md @@ -220,7 +220,7 @@ class Solution: for j in range(coin, amount + 1): dp[j] = min(dp[j], dp[j - coin] + 1) return dp[amount] if dp[amount] < amount + 1 else -1 - + def coinChange1(self, coins: List[int], amount: int) -> int: '''版本二''' # 初始化 @@ -302,6 +302,24 @@ func min(a, b int) int { ``` +Rust: + +```rust +pub fn coin_change(coins: Vec, amount: i32) -> i32 { + let amount = amount as usize; + let mut dp = vec![i32::MAX; amount + 1]; + dp[0] = 0; + for i in 0..coins.len() { + for j in coins[i] as usize..=amount { + if dp[j - coins[i] as usize] != i32::MAX { + dp[j] = dp[j].min(dp[j - coins[i] as usize] + 1); + } + } + } + if dp[amount] == i32::MAX { -1 } else { dp[amount] } +} +``` + Javascript： ```javascript const coinChange = (coins, amount) => { diff --git a/problems/0518.零钱兑换II.md b/problems/0518.零钱兑换II.md index e72c5f85..0e4a3987 100644 --- a/problems/0518.零钱兑换II.md +++ b/problems/0518.零钱兑换II.md @@ -242,6 +242,22 @@ func change(amount int, coins []int) int { } ``` +Rust: +```rust +pub fn change(amount: i32, coins: Vec) -> i32 { + let amount = amount as usize; + let coins = coins.iter().map(|&c|c as usize).collect::>(); + let mut dp = vec![0usize; amount + 1]; + dp[0] = 1; + for i in 0..coins.len() { + for j in coins[i]..=amount { + dp[j] += dp[j - coins[i]]; + } + } + dp[amount] as i32 +} +``` + Javascript： ```javascript const change = (amount, coins) => { diff --git a/problems/0674.最长连续递增序列.md b/problems/0674.最长连续递增序列.md index e941d242..36471490 100644 --- a/problems/0674.最长连续递增序列.md +++ b/problems/0674.最长连续递增序列.md @@ -218,6 +218,7 @@ class Solution: return result ``` + > 贪心法： ```python class Solution: @@ -237,6 +238,24 @@ class Solution: Go： +Rust: +```rust +pub fn find_length_of_lcis(nums: Vec) -> i32 { + if nums.is_empty() { + return 0; + } + let mut result = 1; + let mut dp = vec![1; nums.len()]; + for i in 1..nums.len() { + if nums[i - 1] < nums[i] { + dp[i] = dp[i - 1] + 1; + result = result.max(dp[i]); + } + } + result +} +``` + Javascript： > 动态规划： From c62d518e149295783d219ec1871da84af8e5546e Mon Sep 17 00:00:00 2001 From: FizzerYu <36132150+FizzerYu@users.noreply.github.com> Date: Wed, 11 May 2022 01:17:08 +0800 Subject: [PATCH 080/105] fix bug --- problems/0701.二叉搜索树中的插入操作.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0701.二叉搜索树中的插入操作.md b/problems/0701.二叉搜索树中的插入操作.md index df6a3954..50e39ade 100644 --- a/problems/0701.二叉搜索树中的插入操作.md +++ b/problems/0701.二叉搜索树中的插入操作.md @@ -279,7 +279,7 @@ class Solution: root.right = self.insertIntoBST(root.right, val) # 返回更新后的以当前root为根节点的新树 - return roo + return root ``` **递归法** - 无返回值 From 0e5e657a6d5ba057e8dfa77019b25c5404fe9a9b Mon Sep 17 00:00:00 2001 From: 243wresfdxvc Date: Tue, 10 May 2022 23:04:35 +0000 Subject: [PATCH 081/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200404.=E5=B7=A6?= =?UTF-8?q?=E5=8F=B6=E5=AD=90=E4=B9=8B=E5=92=8C.md=20C=E8=AF=AD=E8=A8=80?= =?UTF-8?q?=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0404.左叶子之和.md | 49 ++++++++++++++++++++++++++++++++ 1 file changed, 49 insertions(+) diff --git a/problems/0404.左叶子之和.md b/problems/0404.左叶子之和.md index 6420da81..d7fd629e 100644 --- a/problems/0404.左叶子之和.md +++ b/problems/0404.左叶子之和.md @@ -466,6 +466,55 @@ func sumOfLeftLeaves(_ root: TreeNode?) -> Int { } ``` +## C +递归法： +```c +int sumOfLeftLeaves(struct TreeNode* root){ + // 递归结束条件：若当前结点为空，返回0 + if(!root) + return 0; + + // 递归取左子树的左结点和和右子树的左结点和 + int leftValue = sumOfLeftLeaves(root->left); + int rightValue = sumOfLeftLeaves(root->right); + + // 若当前结点的左结点存在，且其为叶子结点。取它的值 + int midValue = 0; + if(root->left && (!root->left->left && !root->left->right)) + midValue = root->left->val; + + return leftValue + rightValue + midValue; +} +``` + +迭代法: +```c +int sumOfLeftLeaves(struct TreeNode* root){ + struct TreeNode* stack[1000]; + int stackTop = 0; + + // 若传入root结点不为空，将其入栈 + if(root) + stack[stackTop++] = root; + + int sum = 0; + //若栈不为空，进行循环 + while(stackTop) { + // 出栈栈顶元素 + struct TreeNode *topNode = stack[--stackTop]; + // 若栈顶元素的左孩子为左叶子结点，将其值加入sum中 + if(topNode->left && (!topNode->left->left && !topNode->left->right)) + sum += topNode->left->val; + + // 若当前栈顶结点有左右孩子。将他们加入栈中进行遍历 + if(topNode->right) + stack[stackTop++] = topNode->right; + if(topNode->left) + stack[stackTop++] = topNode->left; + } + return sum; +} +``` ----------------------- From d92aa2c52f0589cca9dbf1a7624f9312395b63fe Mon Sep 17 00:00:00 2001 From: unknown Date: Wed, 11 May 2022 00:39:59 +0100 Subject: [PATCH 082/105] =?UTF-8?q?Add=200112.=E8=B7=AF=E5=BE=84=E6=80=BB?= =?UTF-8?q?=E5=92=8C.md=20C=E8=AF=AD=E8=A8=80=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0112.路径总和.md | 57 +++++++++++++++++++++++++++++++++++ 1 file changed, 57 insertions(+) diff --git a/problems/0112.路径总和.md b/problems/0112.路径总和.md index 41463ec1..6433996c 100644 --- a/problems/0112.路径总和.md +++ b/problems/0112.路径总和.md @@ -1006,6 +1006,63 @@ func traversal(_ cur: TreeNode?, count: Int) { } ``` +## C +0112.路径总和 +递归法： +```c +bool hasPathSum(struct TreeNode* root, int targetSum){ + // 递归结束条件：若当前节点不存在，返回false + if(!root) + return false; + // 若当前节点为叶子节点，且targetSum-root的值为0。（当前路径上的节点值的和满足条件）返回true + if(!root->right && !root->left && targetSum == root->val) + return true; + + // 查看左子树和右子树的所有节点是否满足条件 + return hasPathSum(root->right, targetSum - root->val) || hasPathSum(root->left, targetSum - root->val); +} +``` + +迭代法: +```c +// 存储一个节点以及当前的和 +struct Pair { + struct TreeNode* node; + int sum; +}; + +bool hasPathSum(struct TreeNode* root, int targetSum){ + struct Pair stack[1000]; + int stackTop = 0; + + // 若root存在，则将节点和值封装成一个pair入栈 + if(root) { + struct Pair newPair = {root, root->val}; + stack[stackTop++] = newPair; + } + + // 当栈不为空时 + while(stackTop) { + // 出栈栈顶元素 + struct Pair topPair = stack[--stackTop]; + // 若栈顶元素为叶子节点，且和为targetSum时，返回true + if(!topPair.node->left && !topPair.node->right && topPair.sum == targetSum) + return true; + + // 若当前栈顶节点有左右孩子，计算和并入栈 + if(topPair.node->left) { + struct Pair newPair = {topPair.node->left, topPair.sum + topPair.node->left->val}; + stack[stackTop++] = newPair; + } + if(topPair.node->right) { + struct Pair newPair = {topPair.node->right, topPair.sum + topPair.node->right->val}; + stack[stackTop++] = newPair; + } + } + return false; +} +``` + ----------------------- From cc2c2adb0987a5e2e82eed75a93be068da6388a4 Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Wed, 11 May 2022 16:03:22 +0800 Subject: [PATCH 083/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200977.=E6=9C=89?= =?UTF-8?q?=E5=BA=8F=E6=95=B0=E7=BB=84=E7=9A=84=E5=B9=B3=E6=96=B9.md=20Sca?= =?UTF-8?q?la=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0977.有序数组的平方.md | 34 ++++++++++++++++++++++++++ 1 file changed, 34 insertions(+) diff --git a/problems/0977.有序数组的平方.md b/problems/0977.有序数组的平方.md index 24276bcf..0e79a3d6 100644 --- a/problems/0977.有序数组的平方.md +++ b/problems/0977.有序数组的平方.md @@ -358,7 +358,41 @@ class Solution { } } ``` +Scala: +双指针: +```scala +object Solution { + def sortedSquares(nums: Array[Int]): Array[Int] = { + val res: Array[Int] = new Array[Int](nums.length) + var top = nums.length - 1 + var i = 0 + var j = nums.length - 1 + while (i <= j) { + if (nums(i) * nums(i) <= nums(j) * nums(j)) { + // 当左侧平方小于等于右侧，res数组顶部放右侧的平方，并且top下移，j左移 + res(top) = nums(j) * nums(j) + top -= 1 + j -= 1 + } else { + // 当左侧平方大于右侧，res数组顶部放左侧的平方，并且top下移，i右移 + res(top) = nums(i) * nums(i) + top -= 1 + i += 1 + } + } + res + } +} +``` +骚操作(暴力思路): +```scala +object Solution { + def sortedSquares(nums: Array[Int]): Array[Int] = { + nums.map(x=>{x*x}).sortWith(_ < _) + } +} +``` ----------------------- From 842c04208b163c4782fb7070d949f5feb0c01bee Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Wed, 11 May 2022 16:35:58 +0800 Subject: [PATCH 084/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=88=E8=83=8C?= =?UTF-8?q?=E5=8C=85=E9=97=AE=E9=A2=98=E7=90=86=E8=AE=BA=E5=9F=BA=E7=A1=80?= =?UTF-8?q?=E5=A4=9A=E9=87=8D=E8=83=8C=E5=8C=85.md=EF=BC=89=EF=BC=9A?= =?UTF-8?q?=E5=A2=9E=E5=8A=A0typescript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../背包问题理论基础多重背包.md | 58 +++++++++++++++++++ 1 file changed, 58 insertions(+) diff --git a/problems/背包问题理论基础多重背包.md b/problems/背包问题理论基础多重背包.md index a988db2c..712380f4 100644 --- a/problems/背包问题理论基础多重背包.md +++ b/problems/背包问题理论基础多重背包.md @@ -334,6 +334,64 @@ func Test_multiplePack(t *testing.T) { PASS ``` +TypeScript： + +> 版本一（改变数据源）： + +```typescript +function testMultiPack() { + const bagSize: number = 10; + const weightArr: number[] = [1, 3, 4], + valueArr: number[] = [15, 20, 30], + amountArr: number[] = [2, 3, 2]; + for (let i = 0, length = amountArr.length; i < length; i++) { + while (amountArr[i] > 1) { + weightArr.push(weightArr[i]); + valueArr.push(valueArr[i]); + amountArr[i]--; + } + } + const goodsNum: number = weightArr.length; + const dp: number[] = new Array(bagSize + 1).fill(0); + // 遍历物品 + for (let i = 0; i < goodsNum; i++) { + // 遍历背包容量 + for (let j = bagSize; j >= weightArr[i]; j--) { + dp[j] = Math.max(dp[j], dp[j - weightArr[i]] + valueArr[i]); + } + } + console.log(dp); +} +testMultiPack(); +``` + +> 版本二（改变遍历方式）： + +```typescript +function testMultiPack() { + const bagSize: number = 10; + const weightArr: number[] = [1, 3, 4], + valueArr: number[] = [15, 20, 30], + amountArr: number[] = [2, 3, 2]; + const goodsNum: number = weightArr.length; + const dp: number[] = new Array(bagSize + 1).fill(0); + // 遍历物品 + for (let i = 0; i < goodsNum; i++) { + // 遍历物品个数 + for (let j = 0; j < amountArr[i]; j++) { + // 遍历背包容量 + for (let k = bagSize; k >= weightArr[i]; k--) { + dp[k] = Math.max(dp[k], dp[k - weightArr[i]] + valueArr[i]); + } + } + } + console.log(dp); +} +testMultiPack(); +``` + + + -----------------------
From 871d96a2f9202504f2f5c754da7404f602a9f073 Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Wed, 11 May 2022 16:48:59 +0800 Subject: [PATCH 085/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200209.=E9=95=BF?= =?UTF-8?q?=E5=BA=A6=E6=9C=80=E5=B0=8F=E7=9A=84=E5=AD=90=E6=95=B0=E7=BB=84?= =?UTF-8?q?.md=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0209.长度最小的子数组.md | 48 +++++++++++++++++++++++ 1 file changed, 48 insertions(+) diff --git a/problems/0209.长度最小的子数组.md b/problems/0209.长度最小的子数组.md index fd72cf1b..78b8156c 100644 --- a/problems/0209.长度最小的子数组.md +++ b/problems/0209.长度最小的子数组.md @@ -400,6 +400,54 @@ class Solution { } } ``` +Scala: + +滑动窗口: +```scala +object Solution { + def minSubArrayLen(target: Int, nums: Array[Int]): Int = { + var result = Int.MaxValue // 返回结果，默认最大值 + var left = 0 // 慢指针，当sum>=target，向右移动 + var sum = 0 // 窗口值的总和 + for (right <- 0 until nums.length) { + sum += nums(right) + while (sum >= target) { + result = math.min(result, right - left + 1) // 产生新结果 + sum -= nums(left) // 左指针移动，窗口总和减去左指针的值 + left += 1 // 左指针向右移动 + } + } + // 相当于三元运算符，return关键字可以省略 + if (result == Int.MaxValue) 0 else result + } +} +``` + +暴力解法: +```scala +object Solution { + def minSubArrayLen(target: Int, nums: Array[Int]): Int = { + import scala.util.control.Breaks + var res = Int.MaxValue + var subLength = 0 + for (i <- 0 until nums.length) { + var sum = 0 + Breaks.breakable( + for (j <- i until nums.length) { + sum += nums(j) + if (sum >= target) { + subLength = j - i + 1 + res = math.min(subLength, res) + Breaks.break() + } + } + ) + } + // 相当于三元运算符 + if (res == Int.MaxValue) 0 else res + } +} +``` -----------------------
From cb2fea63e7f5e616f697a715b9d523b6ba877308 Mon Sep 17 00:00:00 2001 From: unknown Date: Wed, 11 May 2022 09:50:47 +0100 Subject: [PATCH 086/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200113.=E8=B7=AF?= =?UTF-8?q?=E5=BE=84=E6=80=BB=E5=92=8CII=20C=E8=AF=AD=E8=A8=80=E8=A7=A3?= =?UTF-8?q?=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0112.路径总和.md | 65 ++++++++++++++++++++++++++++++++++- 1 file changed, 64 insertions(+), 1 deletion(-) diff --git a/problems/0112.路径总和.md b/problems/0112.路径总和.md index 6433996c..de155b45 100644 --- a/problems/0112.路径总和.md +++ b/problems/0112.路径总和.md @@ -1007,7 +1007,7 @@ func traversal(_ cur: TreeNode?, count: Int) { ``` ## C -0112.路径总和 +> 0112.路径总和 递归法： ```c bool hasPathSum(struct TreeNode* root, int targetSum){ @@ -1062,6 +1062,69 @@ bool hasPathSum(struct TreeNode* root, int targetSum){ return false; } ``` +> 0113.路径总和 II +```c +int** ret; +int* path; +int* colSize; +int retTop; +int pathTop; + +void traversal(const struct TreeNode* const node, int count) { + // 若当前节点为叶子节点 + if(!node->right && !node->left) { + // 若当前path上的节点值总和等于targetSum。 + if(count == 0) { + // 复制当前path + int *curPath = (int*)malloc(sizeof(int) * pathTop); + memcpy(curPath, path, sizeof(int) * pathTop); + // 记录当前path的长度为pathTop + colSize[retTop] = pathTop; + // 将当前path加入到ret数组中 + ret[retTop++] = curPath; + } + return; + } + + // 若节点有左/右孩子 + if(node->left) { + // 将左孩子的值加入path中 + path[pathTop++] = node->left->val; + traversal(node->left, count - node->left->val); + // 回溯 + pathTop--; + } + if(node->right) { + // 将右孩子的值加入path中 + path[pathTop++] = node->right->val; + traversal(node->right, count - node->right->val); + // 回溯 + --pathTop; + } +} + +int** pathSum(struct TreeNode* root, int targetSum, int* returnSize, int** returnColumnSizes){ + // 初始化数组 + ret = (int**)malloc(sizeof(int*) * 1000); + path = (int*)malloc(sizeof(int*) * 1000); + colSize = (int*)malloc(sizeof(int) * 1000); + retTop = pathTop = 0; + *returnSize = 0; + + // 若根节点不存在，返回空的ret + if(!root) + return ret; + // 将根节点加入到path中 + path[pathTop++] = root->val; + traversal(root, targetSum - root->val); + + // 设置返回ret数组大小，以及其中每个一维数组元素的长度 + *returnSize = retTop; + *returnColumnSizes = colSize; + + return ret; +} +``` From c363e9da86973e5dafaa5e765c9c6318b5eb9723 Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Wed, 11 May 2022 16:59:05 +0800 Subject: [PATCH 087/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880198.?= =?UTF-8?q?=E6=89=93=E5=AE=B6=E5=8A=AB=E8=88=8D.md=EF=BC=89=EF=BC=9A?= =?UTF-8?q?=E5=A2=9E=E5=8A=A0typescript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0198.打家劫舍.md | 23 +++++++++++++++++++++++ 1 file changed, 23 insertions(+) diff --git a/problems/0198.打家劫舍.md b/problems/0198.打家劫舍.md index dfe1f3a0..a828b9a9 100644 --- a/problems/0198.打家劫舍.md +++ b/problems/0198.打家劫舍.md @@ -189,6 +189,29 @@ const rob = nums => { }; ``` +TypeScript： + +```typescript +function rob(nums: number[]): number { + /** + dp[i]: 前i个房屋能偷到的最大金额 + dp[0]: nums[0]; + dp[1]: max(nums[0], nums[1]); + ... + dp[i]: max(dp[i-1], dp[i-2]+nums[i]); + */ + const length: number = nums.length; + if (length === 1) return nums[0]; + const dp: number[] = []; + dp[0] = nums[0]; + dp[1] = Math.max(nums[0], nums[1]); + for (let i = 2; i < length; i++) { + dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i]); + } + return dp[length - 1]; +}; +``` + From a8cfc460e7143e6b6d8486ad9e8d88e1c9759211 Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Wed, 11 May 2022 17:37:12 +0800 Subject: [PATCH 088/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200059.=E8=9E=BA?= =?UTF-8?q?=E6=97=8B=E7=9F=A9=E9=98=B5II.md=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0059.螺旋矩阵II.md | 51 +++++++++++++++++++++++++++++++++ 1 file changed, 51 insertions(+) diff --git a/problems/0059.螺旋矩阵II.md b/problems/0059.螺旋矩阵II.md index a7b19a34..f3b8e1ce 100644 --- a/problems/0059.螺旋矩阵II.md +++ b/problems/0059.螺旋矩阵II.md @@ -564,6 +564,57 @@ int** generateMatrix(int n, int* returnSize, int** returnColumnSizes){ return ans; } ``` +Scala: +```scala +object Solution { + def generateMatrix(n: Int): Array[Array[Int]] = { + var res = Array.ofDim[Int](n, n) // 定义一个n*n的二维矩阵 + var num = 1 // 标志当前到了哪个数字 + var i = 0 // 横坐标 + var j = 0 // 竖坐标 + while (num <= n * n) { + // 向右：当j不越界，并且下一个要填的数字是空白时 + while (j < n && res(i)(j) == 0) { + res(i)(j) = num // 当前坐标等于num + num += 1 // num++ + j += 1 // 竖坐标+1 + } + i += 1 // 下移一行 + j -= 1 // 左移一列 + + // 剩下的都同上 + + // 向下 + while (i < n && res(i)(j) == 0) { + res(i)(j) = num + num += 1 + i += 1 + } + i -= 1 + j -= 1 + + // 向左 + while (j >= 0 && res(i)(j) == 0) { + res(i)(j) = num + num += 1 + j -= 1 + } + i -= 1 + j += 1 + + // 向上 + while (i >= 0 && res(i)(j) == 0) { + res(i)(j) = num + num += 1 + i -= 1 + } + i += 1 + j += 1 + } + res + } +} +``` -----------------------
From 296485551542d296e27486e985406d7944f3aa9a Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Wed, 11 May 2022 18:35:43 +0800 Subject: [PATCH 089/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880213.?= =?UTF-8?q?=E6=89=93=E5=AE=B6=E5=8A=AB=E8=88=8DII.md=EF=BC=89=EF=BC=9A?= =?UTF-8?q?=E5=A2=9E=E5=8A=A0typescript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0213.打家劫舍II.md | 23 +++++++++++++++++++++++ 1 file changed, 23 insertions(+) diff --git a/problems/0213.打家劫舍II.md b/problems/0213.打家劫舍II.md index 8e569e46..9e698d01 100644 --- a/problems/0213.打家劫舍II.md +++ b/problems/0213.打家劫舍II.md @@ -165,7 +165,30 @@ const robRange = (nums, start, end) => { return dp[end] } ``` +TypeScript： + +```typescript +function rob(nums: number[]): number { + const length: number = nums.length; + if (length === 0) return 0; + if (length === 1) return nums[0]; + return Math.max(robRange(nums, 0, length - 2), + robRange(nums, 1, length - 1)); +}; +function robRange(nums: number[], start: number, end: number): number { + if (start === end) return nums[start]; + const dp: number[] = []; + dp[start] = nums[start]; + dp[start + 1] = Math.max(nums[start], nums[start + 1]); + for (let i = start + 2; i <= end; i++) { + dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i]); + } + return dp[end]; +} +``` + Go： + ```go // 打家劫舍Ⅱ 动态规划 // 时间复杂度O(n) 空间复杂度O(n) From 558dc3343150b1a91a1f307c0bb1fcb881abcd20 Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Wed, 11 May 2022 19:11:53 +0800 Subject: [PATCH 090/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=20=E9=93=BE=E8=A1=A8?= =?UTF-8?q?=E7=90=86=E8=AE=BA=E5=9F=BA=E7=A1=80.md=20Scala=E7=89=88?= =?UTF-8?q?=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/链表理论基础.md | 7 +++++++ 1 file changed, 7 insertions(+) diff --git a/problems/链表理论基础.md b/problems/链表理论基础.md index 095282f5..9dddf1e2 100644 --- a/problems/链表理论基础.md +++ b/problems/链表理论基础.md @@ -210,6 +210,13 @@ type ListNode struct { } ``` +Scala: +```scala +class ListNode(_x: Int = 0, _next: ListNode = null) { + var next: ListNode = _next + var x: Int = _x +} +``` ----------------------- From 8a562b7a2511b44705a6d9e28388738d6a452848 Mon Sep 17 00:00:00 2001 From: UndeadSheep Date: Wed, 11 May 2022 19:44:09 +0800 Subject: [PATCH 091/105] =?UTF-8?q?=E6=96=B0=E5=A2=9E=20=E5=93=88=E5=B8=8C?= =?UTF-8?q?=E8=A1=A8=E9=83=A8=E5=88=86=E7=9A=84=20C#=E7=89=88?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0001.两数之和.md | 18 +++++++++++++++ problems/0202.快乐数.md | 23 +++++++++++++++++++ problems/0242.有效的字母异位词.md | 19 ++++++++++++++++ problems/0349.两个数组的交集.md | 18 +++++++++++++++ problems/0383.赎金信.md | 17 ++++++++++++++ problems/0454.四数相加II.md | 27 +++++++++++++++++++++++ 6 files changed, 122 insertions(+) diff --git a/problems/0001.两数之和.md b/problems/0001.两数之和.md index 9571a773..459a66ad 100644 --- a/problems/0001.两数之和.md +++ b/problems/0001.两数之和.md @@ -274,6 +274,24 @@ class Solution { } } ``` +C#: +```csharp +public class Solution { + public int[] TwoSum(int[] nums, int target) { + Dictionary dic= new Dictionary(); + for(int i=0;i diff --git a/problems/0202.快乐数.md b/problems/0202.快乐数.md index 741a735a..51a79aff 100644 --- a/problems/0202.快乐数.md +++ b/problems/0202.快乐数.md @@ -385,5 +385,28 @@ bool isHappy(int n){ return bHappy; } ``` + +C#： +```csharp +public class Solution { + private int getSum(int n) { + int sum = 0; + //每位数的换算 + while (n > 0) { + sum += (n % 10) * (n % 10); + n /= 10; + } + return sum; + } + public bool IsHappy(int n) { + HashSet set = new HashSet(); + while(n != 1 && !set.Contains(n)) { //判断避免循环 + set.Add(n); + n = getSum(n); + } + return n == 1; + } +} +``` -----------------------
diff --git a/problems/0242.有效的字母异位词.md b/problems/0242.有效的字母异位词.md index 080166fd..878b2466 100644 --- a/problems/0242.有效的字母异位词.md +++ b/problems/0242.有效的字母异位词.md @@ -307,6 +307,25 @@ impl Solution { } } ``` + +C#： +```csharp + public bool IsAnagram(string s, string t) { + int sl=s.Length,tl=t.Length; + if(sl!=tl) return false; + int[] a = new int[26]; + for(int i = 0; i < sl; i++){ + a[s[i] - 'a']++; + a[t[i] - 'a']--; + } + foreach (int i in a) + { + if (i != 0) + return false; + } + return true; + } +``` ## 相关题目 * 383.赎金信 diff --git a/problems/0349.两个数组的交集.md b/problems/0349.两个数组的交集.md index 45f19b6e..7f8958d2 100644 --- a/problems/0349.两个数组的交集.md +++ b/problems/0349.两个数组的交集.md @@ -313,6 +313,24 @@ int* intersection1(int* nums1, int nums1Size, int* nums2, int nums2Size, int* re } ``` +C#: +```csharp + public int[] Intersection(int[] nums1, int[] nums2) { + if(nums1==null||nums1.Length==0||nums2==null||nums1.Length==0) + return new int[0]; //注意数组条件 + HashSet one = Insert(nums1); + HashSet two = Insert(nums2); + one.IntersectWith(two); + return one.ToArray(); + } + public HashSet Insert(int[] nums){ + HashSet one = new HashSet(); + foreach(int num in nums){ + one.Add(num); + } + return one; + } +``` ## 相关题目 * 350.两个数组的交集 II diff --git a/problems/0383.赎金信.md b/problems/0383.赎金信.md index 5d9e8295..6d6efc55 100644 --- a/problems/0383.赎金信.md +++ b/problems/0383.赎金信.md @@ -361,5 +361,22 @@ impl Solution { } ``` +C#： +```csharp +public bool CanConstruct(string ransomNote, string magazine) { + if(ransomNote.Length > magazine.Length) return false; + int[] letters = new int[26]; + foreach(char c in magazine){ + letters[c-'a']++; + } + foreach(char c in ransomNote){ + letters[c-'a']--; + if(letters[c-'a']<0){ + return false; + } + } + return true; + } +``` -----------------------
diff --git a/problems/0454.四数相加II.md b/problems/0454.四数相加II.md index a6cd413b..962fe7a5 100644 --- a/problems/0454.四数相加II.md +++ b/problems/0454.四数相加II.md @@ -318,5 +318,32 @@ impl Solution { } ``` +C#： +```csharp +public int FourSumCount(int[] nums1, int[] nums2, int[] nums3, int[] nums4) { + Dictionary dic = new Dictionary(); + foreach(var i in nums1){ + foreach(var j in nums2){ + int sum = i + j; + if(dic.ContainsKey(sum)){ + dic[sum]++; + }else{ + dic.Add(sum, 1); + } + + } + } + int res = 0; + foreach(var a in nums3){ + foreach(var b in nums4){ + int sum = a+b; + if(dic.TryGetValue(-sum, out var result)){ + res += result; + } + } + } + return res; + } +``` -----------------------
From ef8564a5930805ace91776c5648fa7110800c2d7 Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Wed, 11 May 2022 19:51:07 +0800 Subject: [PATCH 092/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200203.=E7=A7=BB?= =?UTF-8?q?=E9=99=A4=E9=93=BE=E8=A1=A8=E5=85=83=E7=B4=A0.md=20Scala?= =?UTF-8?q?=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0203.移除链表元素.md | 32 ++++++++++++++++++++++++++++- 1 file changed, 31 insertions(+), 1 deletion(-) diff --git a/problems/0203.移除链表元素.md b/problems/0203.移除链表元素.md index c34831b7..53230c0b 100644 --- a/problems/0203.移除链表元素.md +++ b/problems/0203.移除链表元素.md @@ -446,6 +446,36 @@ impl Solution { } } ``` - +Scala: +```scala +/** + * Definition for singly-linked list. + * class ListNode(_x: Int = 0, _next: ListNode = null) { + * var next: ListNode = _next + * var x: Int = _x + * } + */ +object Solution { + def removeElements(head: ListNode, `val`: Int): ListNode = { + if (head == null) return head + var dummy = new ListNode(-1, head) // 定义虚拟头节点 + var cur = head // cur 表示当前节点 + var pre = dummy // pre 表示cur前一个节点 + while (cur != null) { + if (cur.x == `val`) { + // 相等，就删除那么cur的前一个节点pre执行cur的下一个 + pre.next = cur.next + } else { + // 不相等，pre就等于当前cur节点 + pre = cur + } + // 向下迭代 + cur = cur.next + } + // 最终返回dummy的下一个，就是链表的头 + dummy.next + } +} +``` -----------------------
From 898016b8a98aa7eeb2445681e66f0302a79c28ae Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Wed, 11 May 2022 21:13:42 +0800 Subject: [PATCH 093/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200707.=E8=AE=BE?= =?UTF-8?q?=E8=AE=A1=E9=93=BE=E8=A1=A8.md=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0707.设计链表.md | 68 +++++++++++++++++++++++++++++++++++ 1 file changed, 68 insertions(+) diff --git a/problems/0707.设计链表.md b/problems/0707.设计链表.md index 37ce15ad..dcdb53f4 100644 --- a/problems/0707.设计链表.md +++ b/problems/0707.设计链表.md @@ -1154,7 +1154,75 @@ class MyLinkedList { } ``` +Scala: +```scala +class ListNode(_x: Int = 0, _next: ListNode = null) { + var next: ListNode = _next + var x: Int = _x +} +class MyLinkedList() { + + var size = 0 // 链表尺寸 + var dummy: ListNode = new ListNode(0) // 虚拟头节点 + + // 获取第index个节点的值 + def get(index: Int): Int = { + if (index < 0 || index >= size) { + return -1; + } + var cur = dummy + for (i <- 0 to index) { + cur = cur.next + } + cur.x // 返回cur的值 + } + + // 在链表最前面插入一个节点 + def addAtHead(`val`: Int) { + addAtIndex(0, `val`) + } + + // 在链表最后面插入一个节点 + def addAtTail(`val`: Int) { + addAtIndex(size, `val`) + } + + // 在第index个节点之前插入一个新节点 + // 如果index等于链表长度，则说明新插入的节点是尾巴 + // 如果index等于0，则说明新插入的节点是头 + // 如果index>链表长度，则说明为空 + def addAtIndex(index: Int, `val`: Int) { + if (index > size) { + return + } + var loc = index // 因为参数index是val不可变类型，所以需要赋值给一个可变类型 + if (index < 0) { + loc = 0 + } + size += 1 //链表尺寸+1 + var pre = dummy + for (i <- 0 until loc) { + pre = pre.next + } + val node: ListNode = new ListNode(`val`, pre.next) + pre.next = node + } + // 删除第index个节点 + def deleteAtIndex(index: Int) { + if (index < 0 || index >= size) { + return + } + size -= 1 + var pre = dummy + for (i <- 0 until index) { + pre = pre.next + } + pre.next = pre.next.next + } + +} +``` ----------------------- From 923ce563f1461876fbed667f8a79194ccb060096 Mon Sep 17 00:00:00 2001 From: yangtzech Date: Wed, 11 May 2022 21:30:43 +0800 Subject: [PATCH 094/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=20=EF=BC=880102.?= =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=A0=91=E7=9A=84=E5=B1=82=E5=BA=8F=E9=81=8D?= =?UTF-8?q?=E5=8E=86.md=EF=BC=89=EF=BC=9A102.=E4=BA=8C=E5=8F=89=E6=A0=91?= =?UTF-8?q?=E7=9A=84=E5=B1=82=E5=BA=8F=E9=81=8D=E5=8E=86=20=E5=A2=9E?= =?UTF-8?q?=E5=8A=A0=20c++=20=E9=80=92=E5=BD=92=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0102.二叉树的层序遍历.md | 20 ++++++++++++++++++++ 1 file changed, 20 insertions(+) diff --git a/problems/0102.二叉树的层序遍历.md b/problems/0102.二叉树的层序遍历.md index ab8f2e57..5f69f53d 100644 --- a/problems/0102.二叉树的层序遍历.md +++ b/problems/0102.二叉树的层序遍历.md @@ -82,6 +82,26 @@ public: } }; ``` +```CPP +# 递归法 +class Solution { +public: + void order(TreeNode* cur, vector>& result, int depth) + { + if (cur == nullptr) return; + if (result.size() == depth) result.push_back(vector()); + result[depth].push_back(cur->val); + order(cur->left, result, depth + 1); + order(cur->right, result, depth + 1); + } + vector> levelOrder(TreeNode* root) { + vector> result; + int depth = 0; + order(root, result, depth); + return result; + } +}; +``` python3代码： From 3606a624237334eddee98308e39daedbb2d250b0 Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Wed, 11 May 2022 23:03:42 +0800 Subject: [PATCH 095/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200206.=E7=BF=BB?= =?UTF-8?q?=E8=BD=AC=E9=93=BE=E8=A1=A8.md=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0206.翻转链表.md | 34 ++++++++++++++++++++++++++++++++++ 1 file changed, 34 insertions(+) diff --git a/problems/0206.翻转链表.md b/problems/0206.翻转链表.md index 941928ba..25b16907 100644 --- a/problems/0206.翻转链表.md +++ b/problems/0206.翻转链表.md @@ -496,6 +496,40 @@ struct ListNode* reverseList(struct ListNode* head){ return reverse(NULL, head); } ``` +Scala: +双指针法： +```scala +object Solution { + def reverseList(head: ListNode): ListNode = { + var pre: ListNode = null + var cur = head + while (cur != null) { + var tmp = cur.next + cur.next = pre + pre = cur + cur = tmp + } + pre + } +} +``` +递归法： +```scala +object Solution { + def reverseList(head: ListNode): ListNode = { + reverse(null, head) + } + + def reverse(pre: ListNode, cur: ListNode): ListNode = { + if (cur == null) { + return pre // 如果当前cur为空，则返回pre + } + val tmp: ListNode = cur.next + cur.next = pre + reverse(cur, tmp) // 此时cur成为前一个节点，tmp是当前节点 + } +} +``` -----------------------
From 83086c903d77496e6fa34e7e4b0069e27ad1210c Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Thu, 12 May 2022 10:13:00 +0800 Subject: [PATCH 096/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200024.=E4=B8=A4?= =?UTF-8?q?=E4=B8=A4=E4=BA=A4=E6=8D=A2=E9=93=BE=E8=A1=A8=E4=B8=AD=E7=9A=84?= =?UTF-8?q?=E8=8A=82=E7=82=B9.md=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../0024.两两交换链表中的节点.md | 24 ++++++++++++++++++- 1 file changed, 23 insertions(+), 1 deletion(-) diff --git a/problems/0024.两两交换链表中的节点.md b/problems/0024.两两交换链表中的节点.md index ce75e0d7..2289c229 100644 --- a/problems/0024.两两交换链表中的节点.md +++ b/problems/0024.两两交换链表中的节点.md @@ -311,7 +311,29 @@ func swapPairs(_ head: ListNode?) -> ListNode? { return dummyHead.next } ``` - +Scala: +```scala +// 虚拟头节点 +object Solution { + def swapPairs(head: ListNode): ListNode = { + var dummy = new ListNode(0, head) // 虚拟头节点 + var pre = dummy + var cur = head + // 当pre的下一个和下下个都不为空，才进行两两转换 + while (pre.next != null && pre.next.next != null) { + var tmp: ListNode = cur.next.next // 缓存下一次要进行转换的第一个节点 + pre.next = cur.next // 步骤一 + cur.next.next = cur // 步骤二 + cur.next = tmp // 步骤三 + // 下面是准备下一轮的交换 + pre = cur + cur = tmp + } + // 最终返回dummy虚拟头节点的下一个，return可以省略 + dummy.next + } +} +``` -----------------------
From 5fd85215ea9c30c73e5ab84f41dbc9cf170e8617 Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Thu, 12 May 2022 10:53:20 +0800 Subject: [PATCH 097/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200019.=E5=88=A0?= =?UTF-8?q?=E9=99=A4=E9=93=BE=E8=A1=A8=E7=9A=84=E5=80=92=E6=95=B0=E7=AC=AC?= =?UTF-8?q?N=E4=B8=AA=E8=8A=82=E7=82=B9.md=20Scala=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- ...0019.删除链表的倒数第N个节点.md | 24 ++++++++++++++++++- 1 file changed, 23 insertions(+), 1 deletion(-) diff --git a/problems/0019.删除链表的倒数第N个节点.md b/problems/0019.删除链表的倒数第N个节点.md index 813e9b02..8df65627 100644 --- a/problems/0019.删除链表的倒数第N个节点.md +++ b/problems/0019.删除链表的倒数第N个节点.md @@ -289,6 +289,28 @@ func removeNthFromEnd(_ head: ListNode?, _ n: Int) -> ListNode? { return dummyHead.next } ``` - +Scala: +```scala +object Solution { + def removeNthFromEnd(head: ListNode, n: Int): ListNode = { + val dummy = new ListNode(-1, head) // 定义虚拟头节点 + var fast = head // 快指针从头开始走 + var slow = dummy // 慢指针从虚拟头开始头 + // 因为参数 n 是不可变量，所以不能使用 while(n>0){n-=1}的方式 + for (i <- 0 until n) { + fast = fast.next + } + // 快指针和满指针一起走，直到fast走到null + while (fast != null) { + slow = slow.next + fast = fast.next + } + // 删除slow的下一个节点 + slow.next = slow.next.next + // 返回虚拟头节点的下一个 + dummy.next + } +} +``` -----------------------
From 6992735d8de6d1875d0565413faed410e81c822e Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Thu, 12 May 2022 12:46:47 +0800 Subject: [PATCH 098/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=20=E9=9D=A2=E8=AF=95?= =?UTF-8?q?=E9=A2=9802.07.=E9=93=BE=E8=A1=A8=E7=9B=B8=E4=BA=A4.md=20Scala?= =?UTF-8?q?=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/面试题02.07.链表相交.md | 50 ++++++++++++++++++++++++- 1 file changed, 49 insertions(+), 1 deletion(-) diff --git a/problems/面试题02.07.链表相交.md b/problems/面试题02.07.链表相交.md index 2e7226de..0a38cc33 100644 --- a/problems/面试题02.07.链表相交.md +++ b/problems/面试题02.07.链表相交.md @@ -317,7 +317,55 @@ ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) { } ``` - +Scala: +```scala +object Solution { + def getIntersectionNode(headA: ListNode, headB: ListNode): ListNode = { + var lenA = 0 // headA链表的长度 + var lenB = 0 // headB链表的长度 + var tmp = headA // 临时变量 + // 统计headA的长度 + while (tmp != null) { + lenA += 1; + tmp = tmp.next + } + // 统计headB的长度 + tmp = headB // 临时变量赋值给headB + while (tmp != null) { + lenB += 1 + tmp = tmp.next + } + // 因为传递过来的参数是不可变量，所以需要重新定义 + var listA = headA + var listB = headB + // 两个链表的长度差 + // 如果gap>0，lenA>lenB，headA(listA)链表往前移动gap步 + // 如果gap<0，lenA 0) { + // 因为不可以i-=1，所以可以使用for + for (i <- 0 until gap) { + listA = listA.next // 链表headA(listA) 移动 + } + } else { + gap = math.abs(gap) // 此刻gap为负值，取绝对值 + for (i <- 0 until gap) { + listB = listB.next + } + } + // 现在两个链表同时往前走，如果相等则返回 + while (listA != null && listB != null) { + if (listA == listB) { + return listA + } + listA = listA.next + listB = listB.next + } + // 如果链表没有相交则返回null，return可以省略 + null + } +} +``` ----------------------- From 2ce58ac6756775f8ec5139b36a23e8f3de232d30 Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Thu, 12 May 2022 12:55:14 +0800 Subject: [PATCH 099/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880337.?= =?UTF-8?q?=E6=89=93=E5=AE=B6=E5=8A=AB=E8=88=8DIII.md=EF=BC=89=EF=BC=9A?= =?UTF-8?q?=E5=A2=9E=E5=8A=A0typescript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0337.打家劫舍III.md | 43 ++++++++++++++++++++++++++++++++ 1 file changed, 43 insertions(+) diff --git a/problems/0337.打家劫舍III.md b/problems/0337.打家劫舍III.md index a4d8f6b2..6f50723d 100644 --- a/problems/0337.打家劫舍III.md +++ b/problems/0337.打家劫舍III.md @@ -429,7 +429,50 @@ const rob = root => { }; ``` +### TypeScript + +> 记忆化后序遍历 + +```typescript +const memory: Map = new Map(); +function rob(root: TreeNode | null): number { + if (root === null) return 0; + if (memory.has(root)) return memory.get(root); + // 不取当前节点 + const res1: number = rob(root.left) + rob(root.right); + // 取当前节点 + let res2: number = root.val; + if (root.left !== null) res2 += rob(root.left.left) + rob(root.left.right); + if (root.right !== null) res2 += rob(root.right.left) + rob(root.right.right); + const res: number = Math.max(res1, res2); + memory.set(root, res); + return res; +}; +``` + +> 状态标记化后序遍历 + +```typescript +function rob(root: TreeNode | null): number { + return Math.max(...robNode(root)); +}; +// [0]-不偷当前节点能获得的最大金额; [1]-偷~~ +type MaxValueArr = [number, number]; +function robNode(node: TreeNode | null): MaxValueArr { + if (node === null) return [0, 0]; + const leftArr: MaxValueArr = robNode(node.left); + const rightArr: MaxValueArr = robNode(node.right); + // 不偷 + const val1: number = Math.max(leftArr[0], leftArr[1]) + + Math.max(rightArr[0], rightArr[1]); + // 偷 + const val2: number = leftArr[0] + rightArr[0] + node.val; + return [val1, val2]; +} +``` + ### Go + ```go // 打家劫舍Ⅲ 动态规划 // 时间复杂度O(n) 空间复杂度O(logn) From 5da6c06ef9fd308e9ab77b8393e81377eec75821 Mon Sep 17 00:00:00 2001 From: unknown Date: Sat, 14 May 2022 16:27:47 +0100 Subject: [PATCH 100/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=20=E8=83=8C=E5=8C=85?= =?UTF-8?q?=E7=90=86=E8=AE=BA=E8=82=8C=E9=86=8701=E8=83=8C=E5=8C=85=20C?= =?UTF-8?q?=E8=AF=AD=E8=A8=80=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/背包理论基础01背包-1.md | 46 ++++++++++++++++++++++++ 1 file changed, 46 insertions(+) diff --git a/problems/背包理论基础01背包-1.md b/problems/背包理论基础01背包-1.md index fe940b4c..43ad26be 100644 --- a/problems/背包理论基础01背包-1.md +++ b/problems/背包理论基础01背包-1.md @@ -423,5 +423,51 @@ function test () { test(); ``` +### C +```c +#include +#include +#include + +#define MAX(a, b) (((a) > (b)) ? (a) : (b)) +#define ARR_SIZE(a) (sizeof((a)) / sizeof((a)[0])) +#define BAG_WEIGHT 4 + +void backPack(int* weights, int weightSize, int* costs, int costSize, int bagWeight) { + // 开辟dp数组 + int dp[weightSize][bagWeight + 1]; + memset(dp, 0, sizeof(int) * weightSize * (bagWeight + 1)); + + int i, j; + // 当背包容量大于物品0的重量时，将物品0放入到背包中 + for(j = weights[0]; j <= bagWeight; ++j) { + dp[0][j] = costs[0]; + } + + // 先遍历物品，再遍历重量 + for(j = 1; j <= bagWeight; ++j) { + for(i = 1; i < weightSize; ++i) { + // 如果当前背包容量小于物品重量 + if(j < weights[i]) + // 背包物品的价值等于背包不放置当前物品时的价值 + dp[i][j] = dp[i-1][j]; + // 若背包当前重量可以放置物品 + else + // 背包的价值等于放置该物品或不放置该物品的最大值 + dp[i][j] = MAX(dp[i - 1][j], dp[i - 1][j - weights[i]] + costs[i]); + } + } + + printf("%d\n", dp[weightSize - 1][bagWeight]); +} + +int main(int argc, char* argv[]) { + int weights[] = {1, 3, 4}; + int costs[] = {15, 20, 30}; + backPack(weights, ARR_SIZE(weights), costs, ARR_SIZE(costs), BAG_WEIGHT); + return 0; +} +``` + -----------------------
From a5db99363b1bca72345036228776a0320242c8c4 Mon Sep 17 00:00:00 2001 From: SevenMonths Date: Sun, 15 May 2022 15:35:08 +0800 Subject: [PATCH 101/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880035.?= =?UTF-8?q?=E6=90=9C=E7=B4=A2=E6=8F=92=E5=85=A5=E4=BD=8D=E7=BD=AE.md?= =?UTF-8?q?=EF=BC=89=EF=BC=9A=E5=A2=9E=E5=8A=A0PHP=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0035.搜索插入位置.md | 25 +++++++++++++++++++++++++ 1 file changed, 25 insertions(+) diff --git a/problems/0035.搜索插入位置.md b/problems/0035.搜索插入位置.md index 9a770703..8a8f9706 100644 --- a/problems/0035.搜索插入位置.md +++ b/problems/0035.搜索插入位置.md @@ -318,6 +318,31 @@ func searchInsert(_ nums: [Int], _ target: Int) -> Int { ``` +### PHP + +```php +// 二分法(1)：[左闭右闭] +function searchInsert($nums, $target) +{ + $n = count($nums); + $l = 0; + $r = $n - 1; + while ($l <= $r) { + $mid = floor(($l + $r) / 2); + if ($nums[$mid] > $target) { + // 下次搜索在左区间：[$l,$mid-1] + $r = $mid - 1; + } else if ($nums[$mid] < $target) { + // 下次搜索在右区间：[$mid+1,$r] + $l = $mid + 1; + } else { + // 命中返回 + return $mid; + } + } + return $r + 1; +} +``` ----------------------- From aa22b802a5986f06a123c59f395bf49aba779d1a Mon Sep 17 00:00:00 2001 From: unknown Date: Sun, 15 May 2022 14:39:41 +0100 Subject: [PATCH 102/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=20=E8=83=8C=E5=8C=85?= =?UTF-8?q?=E7=90=86=E8=AE=BA=E5=9F=BA=E7=A1=8001-2.mc=20C=E8=AF=AD?= =?UTF-8?q?=E8=A8=80=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/背包理论基础01背包-2.md | 33 ++++++++++++++++++++++++ 1 file changed, 33 insertions(+) diff --git a/problems/背包理论基础01背包-2.md b/problems/背包理论基础01背包-2.md index dabdfb2d..a99a872b 100644 --- a/problems/背包理论基础01背包-2.md +++ b/problems/背包理论基础01背包-2.md @@ -315,6 +315,39 @@ function test () { test(); ``` +### C +```c +#include +#include + +#define MAX(a, b) (((a) > (b)) ? (a) : (b)) +#define ARR_SIZE(arr) ((sizeof((arr))) / sizeof((arr)[0])) +#define BAG_WEIGHT 4 + +void test_back_pack(int* weights, int weightSize, int* values, int valueSize, int bagWeight) { + int dp[bagWeight + 1]; + memset(dp, 0, sizeof(int) * (bagWeight + 1)); + + int i, j; + // 先遍历物品 + for(i = 0; i < weightSize; ++i) { + // 后遍历重量。从后向前遍历 + for(j = bagWeight; j >= weights[i]; --j) { + dp[j] = MAX(dp[j], dp[j - weights[i]] + values[i]); + } + } + + // 打印最优结果 + printf("%d\n", dp[bagWeight]); +} + +int main(int argc, char** argv) { + int weights[] = {1, 3, 4}; + int values[] = {15, 20, 30}; + test_back_pack(weights, ARR_SIZE(weights), values, ARR_SIZE(values), BAG_WEIGHT); + return 0; +} +``` ----------------------- From 19abe18019814af74ee51ba56e4e8ee6106b1253 Mon Sep 17 00:00:00 2001 From: unknown Date: Sun, 15 May 2022 15:26:23 +0100 Subject: [PATCH 103/105] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200416.=E5=88=86?= =?UTF-8?q?=E5=89=B2=E7=AD=89=E5=92=8C=E5=AD=90=E9=9B=86.md=20C=E8=AF=AD?= =?UTF-8?q?=E8=A8=80=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0416.分割等和子集.md | 100 ++++++++++++++++++++++++++++ 1 file changed, 100 insertions(+) diff --git a/problems/0416.分割等和子集.md b/problems/0416.分割等和子集.md index 6e93ae8e..76cfd87f 100644 --- a/problems/0416.分割等和子集.md +++ b/problems/0416.分割等和子集.md @@ -416,6 +416,106 @@ var canPartition = function(nums) { }; ``` +C: +二维dp： +```c +/** +1. dp数组含义：dp[i][j]为背包重量为j时，从[0-i]元素和最大值 +2. 递推公式：dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - nums[i]] + nums[i]) +3. 初始化：dp[i][0]初始化为0。因为背包重量为0时，不可能放入元素。dp[0][j] = nums[0]，当j >= nums[0] && j < target时 +4. 遍历顺序：先遍历物品，再遍历背包 +*/ +#define MAX(a, b) (((a) > (b)) ? (a) : (b)) + +int getSum(int* nums, int numsSize) { + int sum = 0; + + int i; + for(i = 0; i < numsSize; ++i) { + sum += nums[i]; + } + return sum; +} + +bool canPartition(int* nums, int numsSize){ + // 求出元素总和 + int sum = getSum(nums, numsSize); + // 若元素总和为奇数，则不可能得到两个和相等的子数组 + if(sum % 2) + return false; + + // 若子数组的和等于target，则nums可以被分割 + int target = sum / 2; + // 初始化dp数组 + int dp[numsSize][target + 1]; + // dp[j][0]都应被设置为0。因为当背包重量为0时，不可放入元素 + memset(dp, 0, sizeof(int) * numsSize * (target + 1)); + + int i, j; + // 当背包重量j大于nums[0]时，可以在dp[0][j]中放入元素nums[0] + for(j = nums[0]; j <= target; ++j) { + dp[0][j] = nums[0]; + } + + for(i = 1; i < numsSize; ++i) { + for(j = 1; j <= target; ++j) { + // 若当前背包重量j小于nums[i]，则其值等于只考虑0到i-1物品时的值 + if(j < nums[i]) + dp[i][j] = dp[i - 1][j]; + // 否则，背包重量等于在背包中放入num[i]/不放入nums[i]的较大值 + else + dp[i][j] = MAX(dp[i - 1][j], dp[i - 1][j - nums[i]] + nums[i]); + } + } + // 判断背包重量为target，且考虑到所有物品时，放入的元素和是否等于target + return dp[numsSize - 1][target] == target; +} +``` +滚动数组： +```c +/** +1. dp数组含义：dp[j]为背包重量为j时，其中可放入元素的最大值 +2. 递推公式：dp[j] = max(dp[j], dp[j - nums[i]] + nums[i]) +3. 初始化：均初始化为0即可 +4. 遍历顺序：先遍历物品，再后序遍历背包 +*/ +#define MAX(a, b) (((a) > (b)) ? (a) : (b)) + +int getSum(int* nums, int numsSize) { + int sum = 0; + + int i; + for(i = 0; i < numsSize; ++i) { + sum += nums[i]; + } + return sum; +} + +bool canPartition(int* nums, int numsSize){ + // 求出元素总和 + int sum = getSum(nums, numsSize); + // 若元素总和为奇数，则不可能得到两个和相等的子数组 + if(sum % 2) + return false; + // 背包容量 + int target = sum / 2; + + // 初始化dp数组，元素均为0 + int dp[target + 1]; + memset(dp, 0, sizeof(int) * (target + 1)); + + int i, j; + // 先遍历物品，后遍历背包 + for(i = 0; i < numsSize; ++i) { + for(j = target; j >= nums[i]; --j) { + dp[j] = MAX(dp[j], dp[j - nums[i]] + nums[i]); + } + } + + // 查看背包容量为target时，元素总和是否等于target + return dp[target] == target; +} +``` From 83c2fd2454bbbbeda82725d2c38a2b577d7434c6 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=A8=8B=E5=BA=8F=E5=91=98Carl?= Date: Wed, 18 May 2022 09:27:56 +0800 Subject: [PATCH 104/105] Update README.md --- README.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/README.md b/README.md index 3d44ca69..01cab0b0 100644 --- a/README.md +++ b/README.md @@ -31,7 +31,7 @@

-# LeetCode 刷题攻略1111 +# LeetCode 刷题攻略 ## 刷题攻略的背景 From 278011c2df448c295985c0ba4f2ddc0843f4efd7 Mon Sep 17 00:00:00 2001 From: programmercarl <826123027@qq.com> Date: Sun, 22 May 2022 11:11:50 +0800 Subject: [PATCH 105/105] Update --- problems/0056.合并区间.md | 4 +- problems/0077.组合.md | 25 +++++------ problems/0101.对称二叉树.md | 4 +- problems/0435.无重叠区间.md | 2 +- .../0452.用最少数量的箭引爆气球.md | 4 +- problems/面试题 02.07. 解法更新.md | 41 ------------------- 6 files changed, 20 insertions(+), 60 deletions(-) delete mode 100644 problems/面试题 02.07. 解法更新.md diff --git a/problems/0056.合并区间.md b/problems/0056.合并区间.md index a9caeaf0..0d002b46 100644 --- a/problems/0056.合并区间.md +++ b/problems/0056.合并区间.md @@ -112,8 +112,8 @@ public: }; ``` -* 时间复杂度：$O(n\log n)$ ，有一个快排 -* 空间复杂度：$O(1)$，我没有算result数组（返回值所需容器占的空间） +* 时间复杂度：O(nlog n) ，有一个快排 +* 空间复杂度：O(n)，有一个快排，最差情况(倒序)时，需要n次递归调用。因此确实需要O(n)的栈空间 ## 总结 diff --git a/problems/0077.组合.md b/problems/0077.组合.md index 4560c5b7..9e0398ab 100644 --- a/problems/0077.组合.md +++ b/problems/0077.组合.md @@ -27,7 +27,7 @@ 也可以直接看我的B站视频：[带你学透回溯算法-组合问题（对应力扣题目：77.组合）](https://www.bilibili.com/video/BV1ti4y1L7cv#reply3733925949) -# 思路 +## 思路 本题这是回溯法的经典题目。 @@ -232,7 +232,7 @@ void backtracking(参数) { **对比一下本题的代码，是不是发现有点像！** 所以有了这个模板，就有解题的大体方向，不至于毫无头绪。 -# 总结 +## 总结 组合问题是回溯法解决的经典问题，我们开始的时候给大家列举一个很形象的例子，就是n为100，k为50的话，直接想法就需要50层for循环。 @@ -242,7 +242,7 @@ void backtracking(参数) { 接着用回溯法三部曲，逐步分析了函数参数、终止条件和单层搜索的过程。 -# 剪枝优化 +## 剪枝优化 我们说过，回溯法虽然是暴力搜索，但也有时候可以有点剪枝优化一下的。 @@ -324,7 +324,7 @@ public: }; ``` -# 剪枝总结 +## 剪枝总结 本篇我们准对求组合问题的回溯法代码做了剪枝优化，这个优化如果不画图的话，其实不好理解，也不好讲清楚。 @@ -334,10 +334,10 @@ public: -# 其他语言版本 +## 其他语言版本 -## Java： +### Java： ```java class Solution { List> result = new ArrayList<>(); @@ -366,6 +366,8 @@ class Solution { } ``` +### Python + Python2: ```python class Solution(object): @@ -395,7 +397,6 @@ class Solution(object): return result ``` -## Python ```python class Solution: def combine(self, n: int, k: int) -> List[List[int]]: @@ -432,7 +433,7 @@ class Solution: ``` -## javascript +### javascript 剪枝： ```javascript @@ -456,7 +457,7 @@ const combineHelper = (n, k, startIndex) => { } ``` -## TypeScript +### TypeScript ```typescript function combine(n: number, k: number): number[][] { @@ -479,7 +480,7 @@ function combine(n: number, k: number): number[][] { -## Go +### Go ```Go var res [][]int func combine(n int, k int) [][]int { @@ -534,7 +535,7 @@ func backtrack(n,k,start int,track []int){ } ``` -## C +### C ```c int* path; int pathTop; @@ -642,7 +643,7 @@ int** combine(int n, int k, int* returnSize, int** returnColumnSizes){ } ``` -## Swift +### Swift ```swift func combine(_ n: Int, _ k: Int) -> [[Int]] { diff --git a/problems/0101.对称二叉树.md b/problems/0101.对称二叉树.md index e4e232c8..1eb43589 100644 --- a/problems/0101.对称二叉树.md +++ b/problems/0101.对称二叉树.md @@ -238,7 +238,7 @@ public: }; ``` -# 总结 +## 总结 这次我们又深度剖析了一道二叉树的“简单题”，大家会发现，真正的把题目搞清楚其实并不简单，leetcode上accept了和真正掌握了还是有距离的。 @@ -248,7 +248,7 @@ public: 如果已经做过这道题目的同学，读完文章可以再去看看这道题目，思考一下，会有不一样的发现！ -# 相关题目推荐 +## 相关题目推荐 这两道题目基本和本题是一样的，只要稍加修改就可以AC。 diff --git a/problems/0435.无重叠区间.md b/problems/0435.无重叠区间.md index b24ca024..32790a34 100644 --- a/problems/0435.无重叠区间.md +++ b/problems/0435.无重叠区间.md @@ -93,7 +93,7 @@ public: }; ``` * 时间复杂度：O(nlog n) ，有一个快排 -* 空间复杂度：O(1) +* 空间复杂度：O(n)，有一个快排，最差情况(倒序)时，需要n次递归调用。因此确实需要O(n)的栈空间 大家此时会发现如此复杂的一个问题，代码实现却这么简单！ diff --git a/problems/0452.用最少数量的箭引爆气球.md b/problems/0452.用最少数量的箭引爆气球.md index 33bbad55..327694ca 100644 --- a/problems/0452.用最少数量的箭引爆气球.md +++ b/problems/0452.用最少数量的箭引爆气球.md @@ -105,8 +105,8 @@ public: }; ``` -* 时间复杂度：$O(n\log n)$，因为有一个快排 -* 空间复杂度：$O(1)$ +* 时间复杂度：O(nlog n)，因为有一个快排 +* 空间复杂度：O(1)，有一个快排，最差情况(倒序)时，需要n次递归调用。因此确实需要O(n)的栈空间 可以看出代码并不复杂。 diff --git a/problems/面试题 02.07. 解法更新.md b/problems/面试题 02.07. 解法更新.md deleted file mode 100644 index 6115d02e..00000000 --- a/problems/面试题 02.07. 解法更新.md +++ /dev/null @@ -1,41 +0,0 @@ -# 双指针，不计算链表长度 -设置指向headA和headB的指针pa、pb，分别遍历两个链表，每次循环同时更新pa和pb。 -* 当链表A遍历完之后，即pa为空时，将pa指向headB； -* 当链表B遍历完之后，即pa为空时，将pb指向headA； -* 当pa与pb相等时，即指向同一个节点，该节点即为相交起始节点。 -* 若链表不相交，则pa、pb同时为空时退出循环，即如果链表不相交，pa与pb在遍历过全部节点后同时指向结尾空节点，此时退出循环，返回空。 -# 证明思路 -设链表A不相交部分长度为a，链表B不相交部分长度为b，两个链表相交部分长度为c。
-在pa指向链表A时，即pa为空之前，pa经过链表A不相交部分和相交部分，走过的长度为a+c；
-pa指向链表B后，在移动相交节点之前经过链表B不相交部分，走过的长度为b，总合为a+c+b。
-同理，pb走过长度的总合为b+c+a。二者相等，即pa与pb可同时到达相交起始节点。
-该方法可避免计算具体链表长度。 -```cpp -class Solution { -public: - ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) { - //链表为空时，返回空指针 - if(headA == nullptr || headB == nullptr) return nullptr; - ListNode* pa = headA; - ListNode* pb = headB; - //pa与pb在遍历过全部节点后,同时指向结尾空节点时退出循环 - while(pa != nullptr || pb != nullptr){ - //pa为空时，将pa指向headB - if(pa == nullptr){ - pa = headB; - } - //pa为空时，将pb指向headA - if(pb == nullptr){ - pb = headA; - } - //pa与pb相等时，返回相交起始节点 - if(pa == pb){ - return pa; - } - pa = pa->next; - pb = pb->next; - } - return nullptr; - } -}; -```