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 01/72] =?UTF-8?q?Update=200028.=E5=AE=9E=E7=8E=B0strStr.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 02/72] =?UTF-8?q?Update=200459.=E9=87=8D=E5=A4=8D=E7=9A=84?= =?UTF-8?q?=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 03/72] =?UTF-8?q?Update=200018.=E5=9B=9B=E6=95=B0=E4=B9=8B?= =?UTF-8?q?=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 ef44e750cec20980095e63db3c090c8df48f620c Mon Sep 17 00:00:00 2001 From: Guanzhong Pan Date: Mon, 11 Apr 2022 20:34:32 +0100 Subject: [PATCH 04/72] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200376.=E6=91=86?= =?UTF-8?q?=E5=8A=A8=E5=BA=8F=E5=88=97.md=20C=E8=AF=AD=E8=A8=80?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0376.摆动序列.md | 27 +++++++++++++++++++++++++++ 1 file changed, 27 insertions(+) diff --git a/problems/0376.摆动序列.md b/problems/0376.摆动序列.md index 5076c9ad..6bd2277f 100644 --- a/problems/0376.摆动序列.md +++ b/problems/0376.摆动序列.md @@ -298,5 +298,32 @@ var wiggleMaxLength = function(nums) { }; ``` +### C +**贪心** +```c +int wiggleMaxLength(int* nums, int numsSize){ + if(numsSize <= 1) + return numsSize; + + int length = 1; + int preDiff , curDiff; + preDiff = curDiff = 0; + for(int i = 0; i < numsSize - 1; ++i) { + // 计算当前i元素与i+1元素差值 + curDiff = nums[i+1] - nums[i]; + + // 若preDiff与curDiff符号不符，则子序列长度+1。更新preDiff的符号 + // 若preDiff与curDiff符号一致，当前i元素为连续升序/连续降序子序列的中间元素。不被记录入长度 + // 注：当preDiff为0时，curDiff为正或为负都属于符号不同 + if((curDiff > 0 && preDiff <= 0) || (preDiff >= 0 && curDiff < 0)) { + preDiff = curDiff; + length++; + } + } + + return length; +} +``` + -----------------------
From 54d0d9d1c1a579e0a8750f1be864ad78b8d13c45 Mon Sep 17 00:00:00 2001 From: Guanzhong Pan Date: Tue, 12 Apr 2022 10:43:46 +0100 Subject: [PATCH 05/72] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200053.=E6=9C=80?= =?UTF-8?q?=E5=A4=A7=E5=AD=90=E6=95=B0=E7=BB=84=E5=92=8C.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/0053.最大子序和.md | 18 ++++++++++++++++++ 1 file changed, 18 insertions(+) diff --git a/problems/0053.最大子序和.md b/problems/0053.最大子序和.md index 3d11c91e..9af02edf 100644 --- a/problems/0053.最大子序和.md +++ b/problems/0053.最大子序和.md @@ -230,6 +230,24 @@ var maxSubArray = function(nums) { }; ``` +### C: +```c +int maxSubArray(int* nums, int numsSize){ + int maxVal = INT_MIN; + int subArrSum = 0; + + int i; + for(i = 0; i < numsSize; ++i) { + subArrSum += nums[i]; + // 若当前局部和大于之前的最大结果，对结果进行更新 + maxVal = subArrSum > maxVal ? subArrSum : maxVal; + // 若当前局部和为负，对结果无益。则从nums[i+1]开始应重新计算。 + subArrSum = subArrSum < 0 ? 0 : subArrSum; + } + + return maxVal; +} +``` ----------------------- From af724ef6d5ed95c4ba545a932f8257b487648aec Mon Sep 17 00:00:00 2001 From: xuerbujia <83055661+xuerbujia@users.noreply.github.com> Date: Tue, 12 Apr 2022 17:45:50 +0800 Subject: [PATCH 06/72] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880143.=E9=87=8D?= =?UTF-8?q?=E6=8E=92=E9=93=BE=E8=A1=A8.md=EF=BC=89=EF=BC=9A=E5=A2=9E?= =?UTF-8?q?=E5=8A=A0go=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0143.重排链表.md | 28 +++++++++++++++++++++++++++- 1 file changed, 27 insertions(+), 1 deletion(-) diff --git a/problems/0143.重排链表.md b/problems/0143.重排链表.md index 00622623..790bcb48 100644 --- a/problems/0143.重排链表.md +++ b/problems/0143.重排链表.md @@ -336,7 +336,33 @@ class Solution: return pre ``` ### Go - +```go +# 方法三 分割链表 +func reorderList(head *ListNode) { + var slow=head + var fast=head + for fast!=nil&&fast.Next!=nil{ + slow=slow.Next + fast=fast.Next.Next + } //双指针将链表分为左右两部分 + var right =new(ListNode) + for slow!=nil{ + temp:=slow.Next + slow.Next=right.Next + right.Next=slow + slow=temp + } //翻转链表右半部分 + right=right.Next //right为反转后得右半部分 + h:=head + for right.Next!=nil{ + temp:=right.Next + right.Next=h.Next + h.Next=right + h=h.Next.Next + right=temp + } //将左右两部分重新组合 +} +``` ### JavaScript ```javascript From dd63a865717b0a439cc2ce2b5abab0a55d18d7de Mon Sep 17 00:00:00 2001 From: Guanzhong Pan Date: Tue, 12 Apr 2022 11:01:53 +0100 Subject: [PATCH 07/72] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200053.=E6=9C=80?= =?UTF-8?q?=E5=A4=A7=E5=AD=90=E6=95=B0=E7=BB=84=E5=92=8C.md=20C=E8=AF=AD?= =?UTF-8?q?=E8=A8=80=E5=8A=A8=E6=80=81=E8=A7=84=E5=88=92=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0053.最大子序和.md | 34 ++++++++++++++++++++++++++++++++ 1 file changed, 34 insertions(+) diff --git a/problems/0053.最大子序和.md b/problems/0053.最大子序和.md index 9af02edf..3d4459a3 100644 --- a/problems/0053.最大子序和.md +++ b/problems/0053.最大子序和.md @@ -231,6 +231,7 @@ var maxSubArray = function(nums) { ``` ### C: +贪心： ```c int maxSubArray(int* nums, int numsSize){ int maxVal = INT_MIN; @@ -249,6 +250,39 @@ int maxSubArray(int* nums, int numsSize){ } ``` +动态规划： +```c +/** + * 解题思路：动态规划： + * 1. dp数组：dp[i]表示从0到i的子序列中最大序列和的值 + * 2. 递推公式：dp[i] = max(dp[i-1] + nums[i], nums[i]) + 若dp[i-1]<0，对最后结果无益。dp[i]则为nums[i]。 + * 3. dp数组初始化：dp[0]的最大子数组和为nums[0] + * 4. 推导顺序：从前往后遍历 + */ + +#define max(a, b) (((a) > (b)) ? (a) : (b)) + +int maxSubArray(int* nums, int numsSize){ + int dp[numsSize]; + // dp[0]最大子数组和为nums[0] + dp[0] = nums[0]; + // 若numsSize为1，应直接返回nums[0] + int subArrSum = nums[0]; + + int i; + for(i = 1; i < numsSize; ++i) { + dp[i] = max(dp[i - 1] + nums[i], nums[i]); + + // 若dp[i]大于之前记录的最大值，进行更新 + if(dp[i] > subArrSum) + subArrSum = dp[i]; + } + + return subArrSum; +} +``` + -----------------------
From 1b935259a6c83beab9f730059da4bec509bb1f61 Mon Sep 17 00:00:00 2001 From: Guanzhong Pan Date: Tue, 12 Apr 2022 21:01:23 +0100 Subject: [PATCH 08/72] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200122.=E4=B9=B0?= =?UTF-8?q?=E8=82=A1=E7=A5=A8=E7=9A=84=E6=9C=80=E4=BD=B3=E6=97=B6=E6=9C=BA?= =?UTF-8?q?II.md=20C=E8=AF=AD=E8=A8=80=E5=8A=A8=E6=80=81=E8=A7=84=E5=88=92?= =?UTF-8?q?=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../0122.买卖股票的最佳时机II.md | 24 ++++++++++++++++++- 1 file changed, 23 insertions(+), 1 deletion(-) diff --git a/problems/0122.买卖股票的最佳时机II.md b/problems/0122.买卖股票的最佳时机II.md index 83b852c6..1b84f728 100644 --- a/problems/0122.买卖股票的最佳时机II.md +++ b/problems/0122.买卖股票的最佳时机II.md @@ -269,7 +269,7 @@ const maxProfit = (prices) => { ``` C: - +贪心： ```c int maxProfit(int* prices, int pricesSize){ int result = 0; @@ -284,5 +284,27 @@ int maxProfit(int* prices, int pricesSize){ } ``` +动态规划： +```c +#define max(a, b) (((a) > (b)) ? (a) : (b)) + +int maxProfit(int* prices, int pricesSize){ + int dp[pricesSize][2]; + dp[0][0] = 0 - prices[0]; + dp[0][1] = 0; + + int i; + for(i = 1; i < pricesSize; ++i) { + // dp[i][0]为i-1天持股的钱数/在第i天用i-1天的钱买入的最大值。 + // 若i-1天持股，且第i天买入股票比i-1天持股时更亏，说明应在i-1天时持股 + dp[i][0] = max(dp[i-1][0], dp[i-1][1] - prices[i]); + //dp[i][1]为i-1天不持股钱数/在第i天卖出所持股票dp[i-1][0] + prices[i]的最大值 + dp[i][1] = max(dp[i-1][1], dp[i-1][0] + prices[i]); + } + // 返回在最后一天不持股时的钱数（将股票卖出后钱最大化） + return dp[pricesSize - 1][1]; +} +``` + -----------------------
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 09/72] =?UTF-8?q?Update=200459.=E9=87=8D=E5=A4=8D=E7=9A=84?= =?UTF-8?q?=E5=AD=90=E5=AD=97=E7=AC=A6=E4=B8=B2.md=20=E6=B7=BB=E5=8A=A0swi?= =?UTF-8?q?ft=E6=96=B9=E6=B3=95(=E5=89=8D=E7=BC=80=E8=A1=A8=E7=BB=9F?= =?UTF-8?q?=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 10/72] 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 11/72] =?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 12/72] =?UTF-8?q?Update=200459.=E9=87=8D=E5=A4=8D=E7=9A=84?= =?UTF-8?q?=E5=AD=90=E5=AD=97=E7=AC=A6=E4=B8=B2.md=20=E6=B7=BB=E5=8A=A0swi?= =?UTF-8?q?ft=E6=96=B9=E6=B3=95(=E5=89=8D=E7=BC=80=E8=A1=A8=E7=BB=9F?= =?UTF-8?q?=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 e1abedc13ad4d5f744ecc27fa9931aa24508364b Mon Sep 17 00:00:00 2001 From: Guanzhong Pan Date: Wed, 13 Apr 2022 09:41:50 +0100 Subject: [PATCH 13/72] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200055.=E8=B7=B3?= =?UTF-8?q?=E8=B7=83=E6=B8=B8=E6=88=8F.md=20C=E8=AF=AD=E8=A8=80=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/0055.跳跃游戏.md | 22 ++++++++++++++++++++++ 1 file changed, 22 insertions(+) diff --git a/problems/0055.跳跃游戏.md b/problems/0055.跳跃游戏.md index c0890f75..4ba0c03b 100644 --- a/problems/0055.跳跃游戏.md +++ b/problems/0055.跳跃游戏.md @@ -154,6 +154,28 @@ var canJump = function(nums) { }; ``` +### C +```c +#define max(a, b) (((a) > (b)) ? (a) : (b)) + +bool canJump(int* nums, int numsSize){ + int cover = 0; + + int i; + // 只可能获取cover范围中的步数，所以i<=cover + for(i = 0; i <= cover; ++i) { + // 更新cover为从i出发能到达的最大值/cover的值中较大值 + cover = max(i + nums[i], cover); + + // 若更新后cover可以到达最后的元素，返回true + if(cover >= numsSize - 1) + return true; + } + + return false; +} +``` + -----------------------
From b73a7da7585b03b42ab51515f5463259ac2e0631 Mon Sep 17 00:00:00 2001 From: Guanzhong Pan Date: Fri, 15 Apr 2022 18:25:41 +0100 Subject: [PATCH 14/72] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=201005.K=E6=AC=A1?= =?UTF-8?q?=E5=8F=96=E5=8F=8D=E5=90=8E=E6=9C=80=E5=A4=A7=E5=8C=96=E7=9A=84?= =?UTF-8?q?=E6=95=B0=E7=BB=84=E5=92=8C.md=20C=E7=89=88=E6=9C=AC?= 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 | 39 +++++++++++++++++++ 1 file changed, 39 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 45f186e2..0d122d91 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 @@ -211,5 +211,44 @@ var largestSumAfterKNegations = function(nums, k) { }; ``` +### C +```c +#define abs(a) (((a) > 0) ? (a) : (-(a))) + +// 对数组求和 +int sum(int *nums, int numsSize) { + int sum = 0; + + int i; + for(i = 0; i < numsSize; ++i) { + sum += nums[i]; + } + return sum; +} + +int cmp(const void* v1, const void* v2) { + return abs(*(int*)v2) - abs(*(int*)v1); +} + +int largestSumAfterKNegations(int* nums, int numsSize, int k){ + qsort(nums, numsSize, sizeof(int), cmp); + + int i; + for(i = 0; i < numsSize; ++i) { + // 遍历数组，若当前元素<0则将当前元素转变，k-- + if(nums[i] < 0 && k > 0) { + nums[i] *= -1; + --k; + } + } + + // 若遍历完数组后k还有剩余（此时所有元素应均为正），则将绝对值最小的元素nums[numsSize - 1]变为负 + if(k % 2 == 1) + nums[numsSize - 1] *= -1; + + return sum(nums, numsSize); +} +``` + -----------------------
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 15/72] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880209.=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=EF=BC=89=EF=BC=9A=E5=A2=9E=E5=8A=A0Kotlin=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 | 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 16/72] =?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 17/72] =?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 18/72] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880452.=E7=94=A8?= =?UTF-8?q?=E6=9C=80=E5=B0=91=E6=95=B0=E9=87=8F=E7=9A=84=E7=AE=AD=E5=BC=95?= =?UTF-8?q?=E7=88=86=E6=B0=94=E7=90=83.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 --- .../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 84f41f6685e744b138890cf7143ed3069a02a78e Mon Sep 17 00:00:00 2001 From: Guanzhong Pan Date: Tue, 19 Apr 2022 17:00:17 +0100 Subject: [PATCH 19/72] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200135.=E5=88=86?= =?UTF-8?q?=E5=8F=91=E7=B3=96=E6=9E=9C.md=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/0135.分发糖果.md | 41 +++++++++++++++++++++++++++++++++++ 1 file changed, 41 insertions(+) diff --git a/problems/0135.分发糖果.md b/problems/0135.分发糖果.md index ccdabc16..27071aee 100644 --- a/problems/0135.分发糖果.md +++ b/problems/0135.分发糖果.md @@ -238,6 +238,47 @@ var candy = function(ratings) { }; ``` +### C +```c +#define max(a, b) (((a) > (b)) ? (a) : (b)) + +int *initCandyArr(int size) { + int *candyArr = (int*)malloc(sizeof(int) * size); + + int i; + for(i = 0; i < size; ++i) + candyArr[i] = 1; + + return candyArr; +} + +int candy(int* ratings, int ratingsSize){ + // 初始化数组，每个小孩开始至少有一颗糖 + int *candyArr = initCandyArr(ratingsSize); + + int i; + // 先判断右边是否比左边评分高。若是，右边孩子的糖果为左边孩子+1（candyArr[i] = candyArr[i - 1] + 1) + for(i = 1; i < ratingsSize; ++i) { + if(ratings[i] > ratings[i - 1]) + candyArr[i] = candyArr[i - 1] + 1; + } + + // 再判断左边评分是否比右边高。 + // 若是，左边孩子糖果为右边孩子糖果+1/自己所持糖果最大值。（若糖果已经比右孩子+1多，则不需要更多糖果） + // 举例：ratings为[1, 2, 3, 1]。此时评分为3的孩子在判断右边比左边大后为3，虽然它比最末尾的1(ratings[3])大，但是candyArr[3]为1。所以不必更新candyArr[2] + for(i = ratingsSize - 2; i >= 0; --i) { + if(ratings[i] > ratings[i + 1]) + candyArr[i] = max(candyArr[i], candyArr[i + 1] + 1); + } + + // 求出糖果之和 + int result = 0; + for(i = 0; i < ratingsSize; ++i) { + result += candyArr[i]; + } + return result; +} +``` -----------------------
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 20/72] =?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 21/72] =?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 22/72] =?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 23/72] =?UTF-8?q?Update=200309.=E6=9C=80=E4=BD=B3=E4=B9=B0?= =?UTF-8?q?=E5=8D=96=E8=82=A1=E7=A5=A8=E6=97=B6=E6=9C=BA=E5=90=AB=E5=86=B7?= =?UTF-8?q?=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 24/72] =?UTF-8?q?0209.=E9=95=BF=E5=BA=A6=E6=9C=80=E5=B0=8F?= =?UTF-8?q?=E7=9A=84=E5=AD=90=E6=95=B0=E7=BB=84=EF=BC=9A=E8=B0=83=E6=95=B4?= =?UTF-8?q?=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 25/72] =?UTF-8?q?0059.=E8=9E=BA=E6=97=8B=E7=9F=A9=E9=98=B5?= =?UTF-8?q?II=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 26/72] =?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 27/72] =?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 28/72] 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 29/72] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880435.=E6=97=A0?= =?UTF-8?q?=E9=87=8D=E5=8F=A0=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/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 30/72] 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 31/72] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880763.=E5=88=92?= =?UTF-8?q?=E5=88=86=E5=AD=97=E6=AF=8D=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/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 32/72] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880056.=E5=90=88?= =?UTF-8?q?=E5=B9=B6=E5=8C=BA=E9=97=B4.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/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 33/72] =?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 34/72] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880738.=E5=8D=95?= =?UTF-8?q?=E8=B0=83=E9=80=92=E5=A2=9E=E7=9A=84=E6=95=B0=E5=AD=97.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/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 35/72] =?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 36/72] =?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 37/72] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880714.=E4=B9=B0?= =?UTF-8?q?=E5=8D=96=E8=82=A1=E7=A5=A8=E7=9A=84=E6=9C=80=E4=BD=B3=E6=97=B6?= =?UTF-8?q?=E6=9C=BA=E5=90=AB=E6=89=8B=E7=BB=AD=E8=B4=B9.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 --- ...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 38/72] =?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 39/72] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880968.=E7=9B=91?= =?UTF-8?q?=E6=8E=A7=E4=BA=8C=E5=8F=89=E6=A0=91.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/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 40/72] =?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 41/72] =?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 42/72] =?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 43/72] =?UTF-8?q?0300.=E6=9C=80=E9=95=BF=E4=B8=8A=E5=8D=87?= =?UTF-8?q?=E5=AD=90=E5=BA=8F=E5=88=97-go=E5=8A=A8=E6=80=81=E8=A7=84?= =?UTF-8?q?=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 44/72] =?UTF-8?q?0674.=E6=9C=80=E9=95=BF=E8=BF=9E=E7=BB=AD?= =?UTF-8?q?=E9=80=92=E5=A2=9E=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/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 45/72] =?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 46/72] =?UTF-8?q?Update=200106.=E4=BB=8E=E4=B8=AD=E5=BA=8F?= =?UTF-8?q?=E4=B8=8E=E5=90=8E=E5=BA=8F=E9=81=8D=E5=8E=86=E5=BA=8F=E5=88=97?= =?UTF-8?q?=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 47/72] =?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 48/72] =?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 49/72] =?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 50/72] =?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 51/72] =?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 52/72] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880509.=E6=96=90?= =?UTF-8?q?=E6=B3=A2=E9=82=A3=E5=A5=91=E6=95=B0.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/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 53/72] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880070.=E7=88=AC?= =?UTF-8?q?=E6=A5=BC=E6=A2=AF.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.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 54/72] =?UTF-8?q?Update=200383.=E8=B5=8E=E9=87=91=E4=BF=A1?= =?UTF-8?q?.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 55/72] =?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 56/72] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880062.=E4=B8=8D?= =?UTF-8?q?=E5=90=8C=E8=B7=AF=E5=BE=84.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/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 57/72] 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 58/72] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880063.=E4=B8=8D?= =?UTF-8?q?=E5=90=8C=E8=B7=AF=E5=BE=84II.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/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 59/72] =?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 60/72] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880343.=E6=95=B4?= =?UTF-8?q?=E6=95=B0=E6=8B=86=E5=88=86.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/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 61/72] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880096.=E4=B8=8D?= =?UTF-8?q?=E5=90=8C=E7=9A=84=E4=BA=8C=E5=8F=89=E6=90=9C=E7=B4=A2=E6=A0=91?= =?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/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 62/72] =?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 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 63/72] 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 64/72] =?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 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 65/72] =?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 08ae851e75f735f45f033a0c5455aef198d3cc92 Mon Sep 17 00:00:00 2001 From: wang <472146630@qq.com> Date: Fri, 6 May 2022 20:45:29 +0800 Subject: [PATCH 66/72] =?UTF-8?q?=E6=B7=BB=E5=8A=A01035.=E4=B8=8D=E7=9B=B8?= =?UTF-8?q?=E4=BA=A4=E7=9A=84=E7=BA=BF=E5=92=8C1143.=E6=9C=80=E9=95=BF?= =?UTF-8?q?=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 67/72] =?UTF-8?q?Update=200332.=E9=87=8D=E6=96=B0=E5=AE=89?= =?UTF-8?q?=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 68/72] =?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 8bab33d0bba8a080fb614bec6be3b0a5961d7727 Mon Sep 17 00:00:00 2001 From: wang <472146630@qq.com> Date: Tue, 10 May 2022 21:27:54 +0800 Subject: [PATCH 69/72] =?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 0e5e657a6d5ba057e8dfa77019b25c5404fe9a9b Mon Sep 17 00:00:00 2001 From: 243wresfdxvc Date: Tue, 10 May 2022 23:04:35 +0000 Subject: [PATCH 70/72] =?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 a5db99363b1bca72345036228776a0320242c8c4 Mon Sep 17 00:00:00 2001 From: SevenMonths Date: Sun, 15 May 2022 15:35:08 +0800 Subject: [PATCH 71/72] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880035.=E6=90=9C?= =?UTF-8?q?=E7=B4=A2=E6=8F=92=E5=85=A5=E4=BD=8D=E7=BD=AE.md=EF=BC=89?= =?UTF-8?q?=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 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 72/72] 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