From 4fa39a5a47708e9a7cb13cf664acbbfc4a3d5a7b Mon Sep 17 00:00:00 2001 From: "ningwei.shi" Date: Tue, 10 Nov 2020 15:32:02 +0800 Subject: [PATCH 01/63] =?UTF-8?q?=E6=8A=A2=E6=88=BF=E5=AD=90Python3?= =?UTF-8?q?=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 动态规划系列/抢房子.md | 60 +++++++++++++++++++++++++++++++++ 1 file changed, 60 insertions(+) diff --git a/动态规划系列/抢房子.md b/动态规划系列/抢房子.md index 8dae329..ed50060 100644 --- a/动态规划系列/抢房子.md +++ b/动态规划系列/抢房子.md @@ -259,3 +259,63 @@ int[] dp(TreeNode root) {

+[Shantom](https://github.com/Shantom) 提供 198. House Robber I Python3 解法代码： + +```Python +class Solution: + def rob(self, nums: List[int]) -> int: + # 当前，上一间，上上间 + cur, pre1, pre2 = 0, 0, 0 + + for num in nums: + # 当前 = max(上上间+（抢当前），上间（放弃当前）) + cur = max(pre2 + num, pre1) + pre2 = pre1 + pre1 = cur + + return cur +``` +[Shantom](https://github.com/Shantom) 提供 213. House Robber II Python3 解法代码： + +```Python +class Solution: + def rob(self, nums: List[int]) -> int: + # 只有一间时不成环 + if len(nums) == 1: + return nums[0] + + # 该函数同198题 + def subRob(nums: List[int]) -> int: + # 当前，上一间，上上间 + cur, pre1, pre2 = 0, 0, 0 + for num in nums: + # 当前 = max(上上间+（抢当前），上间（放弃当前）) + cur = max(pre2 + num, pre1) + pre2 = pre1 + pre1 = cur + return cur + + # 不考虑第一间或者不考虑最后一间 + return max(subRob(nums[:-1]), subRob(nums[1:])) +``` +[Shantom](https://github.com/Shantom) 提供 337. House Robber III Python3 解法代码： + +```Python +class Solution: + def rob(self, root: TreeNode) -> int: + # 返回值0项为不抢该节点，1项为抢该节点 + def dp(root): + if not root: + return 0, 0 + + left = dp(root.left) + right = dp(root.right) + + do = root.val + left[0] + right[0] + do_not = max(left) + max(right) + + return do_not, do + + return max(dp(root)) +``` + From d0f78c9b7ae615d9af9fc80b8eb05d351a174acb Mon Sep 17 00:00:00 2001 From: Rui Yang <35053274+littlecry@users.noreply.github.com> Date: Tue, 10 Nov 2020 22:57:11 -0600 Subject: [PATCH 02/63] =?UTF-8?q?Update=20=E5=8D=95=E8=B0=83=E6=A0=88.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit add Java solution for Leetcode 739. Daily Temperatures --- 数据结构系列/单调栈.md | 17 ++++++++++++++++- 1 file changed, 16 insertions(+), 1 deletion(-) diff --git a/数据结构系列/单调栈.md b/数据结构系列/单调栈.md index fca0ca8..7853fc1 100644 --- a/数据结构系列/单调栈.md +++ b/数据结构系列/单调栈.md @@ -181,4 +181,19 @@ vector nextGreaterElements(vector& nums) {

-======其他语言代码====== \ No newline at end of file +======其他语言代码====== +// 739. Daily Temperatures +class Solution { + public int[] dailyTemperatures(int[] T) { + Stack stack = new Stack<>(); + int[] ans = new int[T.length]; + for (int i = 0; i < T.length; i++) { + while (!stack.isEmpty() && T[i] > T[stack.peek()]) { + int index = stack.pop(); + ans[index] = i - index; + } + stack.push(i); + } + return ans; + } +} From 91e56f8727f7d5ef1ec5bd1da9944f89592fd196 Mon Sep 17 00:00:00 2001 From: Rui Yang <35053274+littlecry@users.noreply.github.com> Date: Tue, 10 Nov 2020 23:16:02 -0600 Subject: [PATCH 03/63] =?UTF-8?q?Update=20=E5=8D=95=E8=B0=83=E6=A0=88.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit add comments --- 数据结构系列/单调栈.md | 2 ++ 1 file changed, 2 insertions(+) diff --git a/数据结构系列/单调栈.md b/数据结构系列/单调栈.md index 7853fc1..cd68be1 100644 --- a/数据结构系列/单调栈.md +++ b/数据结构系列/单调栈.md @@ -188,8 +188,10 @@ class Solution { Stack stack = new Stack<>(); int[] ans = new int[T.length]; for (int i = 0; i < T.length; i++) { + // 如果压栈之后不满足单调递减，弹出元素，直至保持单调性 while (!stack.isEmpty() && T[i] > T[stack.peek()]) { int index = stack.pop(); + // 被弹出的元素（T[index]）都是小于当前的元素(T[i])，由于栈内元素单调递减，大于被弹出元素（index）的最近的就是当前元素(i) ans[index] = i - index; } stack.push(i); From 93eada462d65a454b3be91c3d1633c0905b4e495 Mon Sep 17 00:00:00 2001 From: MoguCloud Date: Wed, 11 Nov 2020 13:33:41 +0800 Subject: [PATCH 04/63] =?UTF-8?q?=E3=80=9028.=E5=AE=9E=E7=8E=B0=20strStr()?= =?UTF-8?q?=E3=80=91=E3=80=90Python=E3=80=91?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../动态规划之KMP字符匹配算法.md | 38 ++++++++++++++++++- 1 file changed, 37 insertions(+), 1 deletion(-) diff --git a/动态规划系列/动态规划之KMP字符匹配算法.md b/动态规划系列/动态规划之KMP字符匹配算法.md index a0cc4ce..5cde805 100644 --- a/动态规划系列/动态规划之KMP字符匹配算法.md +++ b/动态规划系列/动态规划之KMP字符匹配算法.md @@ -431,4 +431,40 @@ KMP 算法也就是动态规划那点事，我们的公众号文章目录有一

-======其他语言代码====== \ No newline at end of file +======其他语言代码====== +[MoguCloud](https://github.com/MoguCloud) 提供 实现 strStr() 的 Python 完整代码： +```py +class Solution: + def strStr(self, haystack: str, needle: str) -> int: + # 边界条件判断 + if not needle: + return 0 + pat = needle + txt = haystack + + M = len(pat) + # dp[状态][字符] = 下个状态 + dp = [[0 for _ in range(256)] for _ in pat] + # base case + dp[0][ord(pat[0])] = 1 + # 影子状态 X 初始化为 0 + X = 0 + for j in range(1, M): + for c in range(256): + dp[j][c] = dp[X][c] + dp[j][ord(pat[j])] = j + 1 + # 更新影子状态 + X = dp[X][ord(pat[j])] + + N = len(txt) + # pat 初始状态为 0 + j = 0 + for i in range(N): + # 计算 pat 的下一个状态 + j = dp[j][ord(txt[i])] + # 到达终止态，返回结果 + if j == M: + return i - M + 1 + # 没到达终止态，匹配失败 + return -1 +``` From 9606d3437f148a32236fd55342faedb9b40994d6 Mon Sep 17 00:00:00 2001 From: Victor Wu Date: Wed, 11 Nov 2020 13:37:05 +0800 Subject: [PATCH 05/63] =?UTF-8?q?=E4=B8=BA=E8=AE=BE=E8=AE=A1Twitter?= =?UTF-8?q?=E4=B8=80=E6=96=87=EF=BC=8C=E5=AE=9E=E7=8E=B0C++=E7=89=88?= =?UTF-8?q?=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 严格遵守框架。时间戳改成全局变量，并分模块拆开写（参照Java思路） --- 数据结构系列/设计Twitter.md | 119 +++++++++++++++++++++++++++- 1 file changed, 118 insertions(+), 1 deletion(-) diff --git a/数据结构系列/设计Twitter.md b/数据结构系列/设计Twitter.md index 399fbbb..5f6c5f5 100644 --- a/数据结构系列/设计Twitter.md +++ b/数据结构系列/设计Twitter.md @@ -302,4 +302,121 @@ PS：本文前两张图片和 GIF 是我第一次尝试用平板的绘图软件

-======其他语言代码====== \ No newline at end of file +======其他语言代码====== + +[happy-yuxuan](https://github.com/happy-yuxuan) 提供 C++ 代码： + +```c++ +static int timestamp = 0; +class Tweet { +private: + int id; + int time; +public: + Tweet *next; + // id为推文内容，time为发文时间 + Tweet(int id, int time) { + this->id = id; + this->time = time; + next = nullptr; + } + int getId() const { + return this->id; + } + int getTime() const { + return this->time; + } +}; +class User { +private: + int id; +public: + Tweet *head; // 发布的Twitter，用链表表示 + unordered_set followed; // 用户关注了那些人 + User(int userId) { + this->id = userId; + head = nullptr; + // 要先把自己关注了 + followed.insert(id); + } + void follow(int userId) { + followed.insert(userId); + } + void unfollow(int userId) { + // 不可以取关自己 + if (userId != this->id) + followed.erase(userId); + } + void post(int contentId) { + Tweet *twt = new Tweet(contentId, timestamp); + timestamp++; + // 将新建的推文插入链表头 + // 越靠前的推文 timestamp 值越大 + twt->next = head; + head = twt; + } +}; +class Twitter { +private: + // 映射将 userId 和 User 对象对应起来 + unordered_map userMap; + // 判断该用户存不存在系统中,即userMap中存不存在id + inline bool contain(int id) { + return userMap.find(id) != userMap.end(); + } +public: + Twitter() { + userMap.clear(); + } + /* user 发表一条 tweet 动态 */ + void postTweet(int userId, int tweetId) { + if (!contain(userId)) + userMap[userId] = new User(userId); + userMap[userId]->post(tweetId); + } + /* 返回该 user 关注的人（包括他自己）最近的动态 id， + 最多 10 条，而且这些动态必须按从新到旧的时间线顺序排列。*/ + vector getNewsFeed(int userId) { + vector ret; + if (!contain(userId)) return ret; + // 构造一个自动通过Tweet发布的time属性从大到小排序的二叉堆 + typedef function Compare; + Compare cmp = [](const Tweet *a, const Tweet *b) { + return a->getTime() < b->getTime(); + }; + priority_queue, Compare> q(cmp); + // 关注列表的用户Id + unordered_set &users = userMap[userId]->followed; + // 先将所有链表头节点插入优先级队列 + for (int id : users) { + if (!contain(id)) continue; + Tweet *twt = userMap[id]->head; + if (twt == nullptr) continue; + q.push(twt); + } + while (!q.empty()) { + Tweet *t = q.top(); q.pop(); + ret.push_back(t->getId()); + if (ret.size() == 10) return ret; // 最多返回 10 条就够了 + if (t->next) + q.push(t->next); + } + return ret; + } + /* follower 关注 followee */ + void follow(int followerId, int followeeId) { + // 若 follower 不存在，则新建 + if (!contain(followerId)) + userMap[followerId] = new User(followerId); + // 若 followee 不存在，则新建 + if (!contain(followeeId)) + userMap[followeeId] = new User(followeeId); + userMap[followerId]->follow(followeeId); + } + /* follower 取关 followee，如果 Id 不存在则什么都不做 */ + void unfollow(int followerId, int followeeId) { + if (contain(followerId)) + userMap[followerId]->unfollow(followeeId); + } +}; +``` \ No newline at end of file From 0ac247684ae8833de785130a950dd52ca661370e Mon Sep 17 00:00:00 2001 From: tonytang731 <50162066+tonytang731@users.noreply.github.com> Date: Wed, 11 Nov 2020 00:51:19 -0500 Subject: [PATCH 06/63] =?UTF-8?q?[875.=20koko=E5=81=B7=E9=A6=99=E8=95=89][?= =?UTF-8?q?Python3]?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 高频面试系列/koko偷香蕉.md | 41 ++++++++++++++++++++++++++++- 1 file changed, 40 insertions(+), 1 deletion(-) diff --git a/高频面试系列/koko偷香蕉.md b/高频面试系列/koko偷香蕉.md index cc880dd..75109d6 100644 --- a/高频面试系列/koko偷香蕉.md +++ b/高频面试系列/koko偷香蕉.md @@ -169,4 +169,43 @@ for (int i = 0; i < n; i++)

-======其他语言代码====== \ No newline at end of file +======其他语言代码====== +```python +import math + +class Solution: + def minEatingSpeed(self, piles, H): + # 初始化起点和终点， 最快的速度可以一次拿完最大的一堆 + start = 1 + end = max(piles) + + # while loop进行二分查找 + while start + 1 < end: + mid = start + (end - start) // 2 + + # 如果中点所需时间大于H, 我们需要加速， 将起点设为中点 + if self.timeH(piles, mid) > H: + start = mid + # 如果中点所需时间小于H, 我们需要减速， 将终点设为中点 + else: + end = mid + + # 提交前确认起点是否满足条件，我们要尽量慢拿 + if self.timeH(piles, start) <= H: + return start + + # 若起点不符合， 则中点是答案 + return end + + + + def timeH(self, piles, K): + # 初始化时间 + H = 0 + + #求拿每一堆需要多长时间 + for pile in piles: + H += math.ceil(pile / K) + + return H +``` From 4870bf9504542870f41e2762cab28785158229d1 Mon Sep 17 00:00:00 2001 From: Victor Wu Date: Wed, 11 Nov 2020 13:52:53 +0800 Subject: [PATCH 07/63] =?UTF-8?q?=E8=A1=A5=E5=85=85Java=E4=BB=A3=E7=A0=81?= =?UTF-8?q?=E4=B8=AD=E6=BC=8F=E6=8E=89=E4=BA=86=E4=B8=80=E4=B8=AA=E5=A4=A7?= =?UTF-8?q?=E6=8B=AC=E5=8F=B7?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 高频面试系列/消失的元素.md | 1 + 1 file changed, 1 insertion(+) diff --git a/高频面试系列/消失的元素.md b/高频面试系列/消失的元素.md index 844b5cb..5d65c2a 100644 --- a/高频面试系列/消失的元素.md +++ b/高频面试系列/消失的元素.md @@ -89,6 +89,7 @@ int missingNumber(int[] nums) { for (int x : nums) sum += x; return expect - sum; +} ``` 你看，这种解法应该是最简单的，但说实话，我自己也没想到这个解法，而且我去问了几个大佬，他们也没想到这个最简单的思路。相反，如果去问一个初中生，他也许很快就能想到。 From c1237d7155658f77b7f26426ca36b5ee9b09caa6 Mon Sep 17 00:00:00 2001 From: tonytang731 <50162066+tonytang731@users.noreply.github.com> Date: Wed, 11 Nov 2020 01:06:53 -0500 Subject: [PATCH 08/63] =?UTF-8?q?Update=20koko=E5=81=B7=E9=A6=99=E8=95=89.?= =?UTF-8?q?md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 高频面试系列/koko偷香蕉.md | 2 ++ 1 file changed, 2 insertions(+) diff --git a/高频面试系列/koko偷香蕉.md b/高频面试系列/koko偷香蕉.md index 75109d6..635291c 100644 --- a/高频面试系列/koko偷香蕉.md +++ b/高频面试系列/koko偷香蕉.md @@ -170,6 +170,8 @@ for (int i = 0; i < n; i++)

======其他语言代码====== + +[tonytang731](https://https://github.com/tonytang731) 提供 Python3 代码： ```python import math From cd51fcf2b62b7a242d3d21949254e1a233207d76 Mon Sep 17 00:00:00 2001 From: Victor Wu Date: Wed, 11 Nov 2020 14:02:23 +0800 Subject: [PATCH 09/63] =?UTF-8?q?=E3=80=90268.=20=E4=B8=A2=E5=A4=B1?= =?UTF-8?q?=E7=9A=84=E6=95=B0=E5=AD=97=E3=80=91=E3=80=90C++=E3=80=91?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 高频面试系列/消失的元素.md | 47 ++++++++++++++++++++++++++- 1 file changed, 46 insertions(+), 1 deletion(-) diff --git a/高频面试系列/消失的元素.md b/高频面试系列/消失的元素.md index 844b5cb..50556ba 100644 --- a/高频面试系列/消失的元素.md +++ b/高频面试系列/消失的元素.md @@ -132,4 +132,49 @@ public int missingNumber(int[] nums) {

-======其他语言代码====== \ No newline at end of file +======其他语言代码====== + +[happy-yuxuan](https://github.com/happy-yuxuan) 提供 三种方法的 C++ 代码： + +```c++ +// 方法：异或元素和索引 +int missingNumber(vector& nums) { + int n = nums.size(); + int res = 0; + // 先和新补的索引异或一下 + res ^= n; + // 和其他的元素、索引做异或 + for (int i = 0; i < n; i++) + res ^= i ^ nums[i]; + return res; +} +``` + +```c++ +// 方法：等差数列求和 +int missingNumber(vector& nums) { + int n = nums.size(); + // 公式：(首项 + 末项) * 项数 / 2 + int expect = (0 + n) * (n + 1) / 2; + int sum = 0; + for (int x : nums) + sum += x; + return expect - sum; +} +``` + +```c++ +// 方法：防止整型溢出 +int missingNumber(vector& nums) { + int n = nums.size(); + int res = 0; + // 新补的索引 + res += n - 0; + // 剩下索引和元素的差加起来 + for (int i = 0; i < n; i++) + res += i - nums[i]; + return res; +} +``` + + From 88fe014eb64275d4909f7f27fa03fd2246dbdca3 Mon Sep 17 00:00:00 2001 From: Justlxb0124 <33374844+Justlxb0124@users.noreply.github.com> Date: Wed, 11 Nov 2020 14:23:01 +0800 Subject: [PATCH 10/63] =?UTF-8?q?Update=20=E5=8D=95=E8=B0=83=E9=98=9F?= =?UTF-8?q?=E5=88=97.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 数据结构系列/单调队列.md | 49 +++++++++++++++++++++++++++++- 1 file changed, 48 insertions(+), 1 deletion(-) diff --git a/数据结构系列/单调队列.md b/数据结构系列/单调队列.md index eb298a1..d82dbe0 100644 --- a/数据结构系列/单调队列.md +++ b/数据结构系列/单调队列.md @@ -210,4 +210,51 @@ vector maxSlidingWindow(vector& nums, int k) {

-======其他语言代码====== \ No newline at end of file +======其他语言代码====== +```java +class Solution { + public int[] maxSlidingWindow(int[] nums, int k) { + int len = nums.length; + // 判断数组或者窗口长度为0的情况 + if (len * k == 0) { + return new int[0]; + } + + /* + 采用两端扫描的方法 + 将数组分成大小为 k 的若干个窗口, 对每个窗口分别从左往右和从右往左扫描, 记录扫描的最大值 + left[] 记录从左往右扫描的最大值 + right[] 记录从右往左扫描的最大值 + */ + int[] left = new int[len]; + int[] right = new int[len]; + + for (int i = 0; i < len; i = i + k) { + // 每个窗口中的第一个值 + left[i] = nums[i]; + // 窗口的最后边界 + int index = i + k - 1 >= len ? len - 1 : i + k - 1; + // 每个窗口的最后一个值 + right[index] = nums[index]; + // 对该窗口从左往右扫描 + for (int j = i + 1; j <= index; j++) { + left[j] = Math.max(left[j - 1], nums[j]); + } + // 对该窗口从右往左扫描 + for (int j = index - 1; j >= i; j--) { + right[j] = Math.max(right[j + 1], nums[j]); + } + } + + int[] arr = new int[len - k + 1]; + + // 对于第 i 个位置, 它一定是该窗口从右往左扫描数组中的最后一个值, 相对的 i + k - 1 是该窗口从左向右扫描数组中的最后一个位置 + // 对两者取最大值即可 + for (int i = 0; i < len - k + 1; i++) { + arr[i] = Math.max(right[i], left[i + k - 1]); + } + + return arr; + } +} +``` From a72173b1ef5c3660b47b8364ca3e19ee5a7b7478 Mon Sep 17 00:00:00 2001 From: dekunma <53892579+dekunma@users.noreply.github.com> Date: Wed, 11 Nov 2020 14:23:51 +0800 Subject: [PATCH 11/63] =?UTF-8?q?=E3=80=90392.=20=E5=88=A4=E6=96=AD?= =?UTF-8?q?=E5=AD=90=E5=BA=8F=E5=88=97=E3=80=91=E3=80=90C++=E3=80=91?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../二分查找判定子序列.md | 65 ++++++++++++++++++- 1 file changed, 64 insertions(+), 1 deletion(-) diff --git a/高频面试系列/二分查找判定子序列.md b/高频面试系列/二分查找判定子序列.md index c71e65f..e9c5b0f 100644 --- a/高频面试系列/二分查找判定子序列.md +++ b/高频面试系列/二分查找判定子序列.md @@ -168,4 +168,67 @@ boolean isSubsequence(String s, String t) {

-======其他语言代码====== \ No newline at end of file +======其他语言代码====== +[dekunma](https://www.linkedin.com/in/dekun-ma-036a9b198/) 提供C++代码 +**解法一：遍历（也可以用双指针）：** +```C++ +class Solution { +public: + bool isSubsequence(string s, string t) { + // 遍历s + for(int i = 0; i < s.size(); i++) { + // 找到s[i]字符在t中的位置 + size_t pos = t.find(s[i]); + + // 如果s[i]字符不在t中，返回false + if(pos == std::string::npos) return false; + // 如果s[i]在t中，后面就只看pos以后的字串，防止重复查找 + else t = t.substr(pos + 1); + } + return true; + } +}; +``` + +**解法二：二分查找：** +```C++ +class Solution { +public: + bool isSubsequence(string s, string t) { + int m = s.size(), n = t.size(); + // 对 t 进行预处理 + vector index[256]; + for (int i = 0; i < n; i++) { + char c = t[i]; + index[c].push_back(i); + } + // 串 t 上的指针 + int j = 0; + // 借助 index 查找 s[i] + for (int i = 0; i < m; i++) { + char c = s[i]; + // 整个 t 压根儿没有字符 c + if (index[c].empty()) return false; + int pos = left_bound(index[c], j); + // 二分搜索区间中没有找到字符 c + if (pos == index[c].size()) return false; + // 向前移动指针 j + j = index[c][pos] + 1; + } + return true; + } + // 查找左侧边界的二分查找 + int left_bound(vector arr, int tar) { + int lo = 0, hi = arr.size(); + while (lo < hi) { + int mid = lo + (hi - lo) / 2; + if (tar > arr[mid]) { + lo = mid + 1; + } else { + hi = mid; + } + } + return lo; + } +}; +``` From 1c871a9eb0d5a556b41409ac86750b878dd72e17 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E6=9C=B1=E5=8A=9B?= <171250568@smail.nju.edu.cn> Date: Wed, 11 Nov 2020 14:41:25 +0800 Subject: [PATCH 12/63] =?UTF-8?q?=E3=80=90645.=20=E9=94=99=E8=AF=AF?= =?UTF-8?q?=E7=9A=84=E9=9B=86=E5=90=88=E3=80=91=E3=80=90Java=E3=80=91?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../缺失和重复的元素.md | 30 +++++++++++++++++-- 1 file changed, 28 insertions(+), 2 deletions(-) diff --git a/高频面试系列/缺失和重复的元素.md b/高频面试系列/缺失和重复的元素.md index 06e554f..52c8d78 100644 --- a/高频面试系列/缺失和重复的元素.md +++ b/高频面试系列/缺失和重复的元素.md @@ -1,4 +1,4 @@ -# 如何寻找缺失和重复的元素 +# 如何寻找缺失和重复的元素

@@ -139,4 +139,30 @@ vector findErrorNums(vector& nums) {

-======其他语言代码====== \ No newline at end of file +======其他语言代码====== + +[zhuli](https://github.com/1097452462 "zhuli")提供的Java代码： +```java +class Solution { + public int[] findErrorNums(int[] nums) { + int n = nums.length; + int dup = -1; + for (int i = 0; i < n; i++) { + // 元素是从 1 开始的 + int index = Math.abs(nums[i]) - 1; + // nums[index] 小于 0 则说明重复访问 + if (nums[index] < 0) + dup = Math.abs(nums[i]); + else + nums[index] *= -1; + } + int missing = -1; + for (int i = 0; i < n; i++) + // nums[i] 大于 0 则说明没有访问 + if (nums[i] > 0) + // 将索引转换成元素 + missing = i + 1; + return new int[]{dup, missing}; + } +} +``` \ No newline at end of file From ac37263996303704f162df98bf90358c426bfbef Mon Sep 17 00:00:00 2001 From: Chenjie Xu Date: Wed, 11 Nov 2020 15:07:41 +0800 Subject: [PATCH 13/63] =?UTF-8?q?=E3=80=9098.=20=E9=AA=8C=E8=AF=81?= =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=90=9C=E7=B4=A2=E6=A0=91=E3=80=91=E3=80=90?= =?UTF-8?q?Python=E3=80=91?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../二叉搜索树操作集锦.md | 23 +++++++++++++++++++ 1 file changed, 23 insertions(+) diff --git a/数据结构系列/二叉搜索树操作集锦.md b/数据结构系列/二叉搜索树操作集锦.md index 6c05626..8281c20 100644 --- a/数据结构系列/二叉搜索树操作集锦.md +++ b/数据结构系列/二叉搜索树操作集锦.md @@ -309,3 +309,26 @@ void BST(TreeNode root, int target) {

+ +[ChenjieXu](https://github.com/ChenjieXu)提供第98题Python3代码： + +```python +def isValidBST(self, root): + # 递归函数 + def helper(node, lower = float('-inf'), upper = float('inf')): + if not node: + return True + + val = node.val + if val <= lower or val >= upper: + return False + # 右节点 + if not helper(node.right, val, upper): + return False + # 左节点 + if not helper(node.left, lower, val): + return False + return True + + return helper(root) +``` From 65e137edb75ba5bcbfdf571155c6cf6cbc11863b Mon Sep 17 00:00:00 2001 From: Volvane Date: Wed, 11 Nov 2020 15:26:31 +0800 Subject: [PATCH 14/63] =?UTF-8?q?Update=20LRU=E7=AE=97=E6=B3=95.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 146题C++代码 --- 高频面试系列/LRU算法.md | 88 ++++++++++++++++++++++++++++++++- 1 file changed, 87 insertions(+), 1 deletion(-) diff --git a/高频面试系列/LRU算法.md b/高频面试系列/LRU算法.md index b9ca311..23e0b65 100644 --- a/高频面试系列/LRU算法.md +++ b/高频面试系列/LRU算法.md @@ -346,4 +346,90 @@ class LRUCache {

-======其他语言代码====== \ No newline at end of file +======其他语言代码====== + +[gowufang](https://github.com/gowufang)提供第146题C++代码： +```cpp +class LRUCache { + public: + struct node { + int val; + int key; + node* pre;//当前节点的前一个节点 + node* next;//当前节点的后一个节点 + node(){} + node(int key, int val):key(key), val(val), pre(NULL), next(NULL){} + }; + + LRUCache(int size) { + this->size = size; + head = new node(); + tail = new node(); + head->next = tail; + tail->pre = head; + } + + + void movetohead(node* cur)//相当于一个insert操作，在head 和 head的next之间插入一个节点 + { + node* next = head->next;//head的next先保存起来 + head->next = cur;//将当前节点移动到head的后面 + cur->pre = head;//当前节点cur的pre指向head + next->pre = cur; + cur->next = next; + } + + node* deletecurrentnode(node* cur)//移除当前节点 + { + cur->pre->next = cur->next; + cur->next->pre = cur->pre; + return cur; + } + void makerecently(node* cur) + { + node* temp = deletecurrentnode(cur);// 删除 cur，要重新插入到对头 + movetohead(temp);//cur放到队头去 + } + int get(int key) + { + int ret = -1; + if ( map.count(key)) + { + node* temp = map[key]; + makerecently(temp);// 将 key 变为最近使用 + ret = temp->val; + } + return ret; + } + + void put(int key, int value) { + if ( map.count(key)) + { + // 修改 key 的值 + node* temp = map[key]; + temp->val = value; + // 将 key 变为最近使用 + makerecently(temp); + } + else + { + node* cur = new node(key, value); + if( map.size()== size ) + { + // 链表头部就是最久未使用的 key + node *temp = deletecurrentnode(tail->pre); + map.erase(temp->key); + } + movetohead(cur); + map[key] = cur; + + } + + } + + unordered_map map; + int size; + node* head, *tail; + + }; + ``` From 022fe8afccac97df3b5a071799e032a3f1b935f0 Mon Sep 17 00:00:00 2001 From: lixiandea Date: Wed, 11 Nov 2020 15:37:34 +0800 Subject: [PATCH 15/63] =?UTF-8?q?=E3=80=90100.=20=E7=9B=B8=E5=90=8C?= =?UTF-8?q?=E7=9A=84=E6=A0=91=E3=80=91=E3=80=90Python=E3=80=91?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../二叉搜索树操作集锦.md | 27 ++++++++++++++++++- 1 file changed, 26 insertions(+), 1 deletion(-) diff --git a/数据结构系列/二叉搜索树操作集锦.md b/数据结构系列/二叉搜索树操作集锦.md index 801b8fb..45b77df 100644 --- a/数据结构系列/二叉搜索树操作集锦.md +++ b/数据结构系列/二叉搜索树操作集锦.md @@ -310,4 +310,29 @@ void BST(TreeNode root, int target) {

-======其他语言代码====== \ No newline at end of file +======其他语言代码====== + +[lixiandea](https://github.com/lixiandea)提供第100题Python3代码： +```python3 +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right +class Solution: + def isSameTree(self, p: TreeNode, q: TreeNode) -> bool: + ''' + 当前节点值相等且树的子树相等，则树相等。 + 递归退出条件：两个节点存在一个节点为空 + ''' + if p == None: + if q == None: + return True + else: + return False + if q == None: + return False + # 当前节点相同且左子树和右子树分别相同 + return p.val==q.val and self.isSameTree(p.left, q.left) and self.isSameTree(p.right, q.right) +``` \ No newline at end of file From ce67219141fbe62377facafe1024a81a3ad3eb85 Mon Sep 17 00:00:00 2001 From: hzs Date: Wed, 11 Nov 2020 16:03:43 +0800 Subject: [PATCH 16/63] =?UTF-8?q?[239.=E6=BB=91=E5=8A=A8=E7=AA=97=E5=8F=A3?= =?UTF-8?q?=E6=9C=80=E5=A4=A7=E5=80=BC]=20[Python]?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 数据结构系列/单调队列.md | 50 +++++++++++++++++++++++++++++- 1 file changed, 49 insertions(+), 1 deletion(-) diff --git a/数据结构系列/单调队列.md b/数据结构系列/单调队列.md index eb298a1..41b494f 100644 --- a/数据结构系列/单调队列.md +++ b/数据结构系列/单调队列.md @@ -210,4 +210,52 @@ vector maxSlidingWindow(vector& nums, int k) {

-======其他语言代码====== \ No newline at end of file +======其他语言代码====== + +python3版本 + +```python +from collections import deque + +class MonotonicQueue(object): + def __init__(self): + # 双端队列 + self.data = deque() + + def push(self, n): + # 实现单调队列的push方法 + while self.data and self.data[-1] < n: + self.data.pop() + self.data.append(n) + + def max(self): + # 取得单调队列中的最大值 + return self.data[0] + + def pop(self, n): + # 实现单调队列的pop方法 + if self.data and self.data[0] == n: + self.data.popleft() + + +class Solution: + def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]: + # 单调队列实现的窗口 + window = MonotonicQueue() + + # 结果 + res = [] + + for i in range(0, len(nums)): + + if i < k-1: + # 先填满窗口前k-1 + window.push(nums[i]) + else: + # 窗口向前滑动 + window.push(nums[i]) + res.append(window.max()) + window.pop(nums[i-k+1]) + return res + +``` \ No newline at end of file From 42845993b606dc6d3b54bcba29e37ee8d26dda48 Mon Sep 17 00:00:00 2001 From: Andrew <920076768@qq.com> Date: Wed, 11 Nov 2020 16:18:15 +0800 Subject: [PATCH 17/63] =?UTF-8?q?Update=20=E9=80=92=E5=BD=92=E5=8F=8D?= =?UTF-8?q?=E8=BD=AC=E9=93=BE=E8=A1=A8=E7=9A=84=E4=B8=80=E9=83=A8=E5=88=86?= =?UTF-8?q?.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 增加了python版本代码 --- .../递归反转链表的一部分.md | 38 ++++++++++++++++++- 1 file changed, 37 insertions(+), 1 deletion(-) diff --git a/数据结构系列/递归反转链表的一部分.md b/数据结构系列/递归反转链表的一部分.md index 830e503..1f8bca2 100644 --- a/数据结构系列/递归反转链表的一部分.md +++ b/数据结构系列/递归反转链表的一部分.md @@ -218,4 +218,40 @@ ListNode reverseBetween(ListNode head, int m, int n) {

-======其他语言代码====== \ No newline at end of file +======其他语言代码====== + +[DiamondI](https://github.com/DiamondI) 提供python3版本代码： + +思路：递归。时间复杂度为O(n)，由于递归调用需要借助栈的空间，因此空间复杂度亦为O(n)。 + +```python3 +# Definition for singly-linked list. +# class ListNode: +# def __init__(self, val=0, next=None): +# self.val = val +# self.next = next +class Solution: + def __init__(self): + self.__successor = None + + def __reverseN(self, head: ListNode, n: int) -> ListNode: + if n == 1: + # 记录第 n + 1 个节点 + self.__successor = head.next; + return head; + # 以 head.next 为起点，需要反转前 n - 1 个节点 + last = self.__reverseN(head.next, n - 1); + + head.next.next = head; + # 让反转之后的 head 节点和后面的节点连起来 + head.next = self.__successor; + return last; + + def reverseBetween(self, head: ListNode, m: int, n: int) -> ListNode: + # base case + if m == 1: + return self.__reverseN(head, n); + # 前进到反转的起点触发 base case + head.next = self.reverseBetween(head.next, m - 1, n - 1); + return head; +``` From c786b6ab43c7c24e55b4877ea6c1afafae1a2179 Mon Sep 17 00:00:00 2001 From: hzs Date: Wed, 11 Nov 2020 16:19:53 +0800 Subject: [PATCH 18/63] =?UTF-8?q?=E3=80=90239.=E6=BB=91=E5=8A=A8=E7=AA=97?= =?UTF-8?q?=E5=8F=A3=E6=9C=80=E5=A4=A7=E5=80=BC=E3=80=91=E3=80=90Python?= =?UTF-8?q?=E3=80=91?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 数据结构系列/单调队列.md | 4 ++++ 1 file changed, 4 insertions(+) diff --git a/数据结构系列/单调队列.md b/数据结构系列/单调队列.md index 41b494f..094e510 100644 --- a/数据结构系列/单调队列.md +++ b/数据结构系列/单调队列.md @@ -210,10 +210,14 @@ vector maxSlidingWindow(vector& nums, int k) {

+ ======其他语言代码====== python3版本 +由[SCUHZS](ttps://github.com/brucecat)提供 + + ```python from collections import deque From 0a4bddfe2b3a540476920d50c17c6457e86344be Mon Sep 17 00:00:00 2001 From: forthespada <44971298+forthespada@users.noreply.github.com> Date: Wed, 11 Nov 2020 16:21:27 +0800 Subject: [PATCH 19/63] =?UTF-8?q?=E3=80=90234.=E6=BB=91=E5=8A=A8=E7=AA=97?= =?UTF-8?q?=E5=8F=A3=E6=9C=80=E5=A4=A7=E5=80=BC=E3=80=91=E3=80=90=E5=9B=9E?= =?UTF-8?q?=E6=96=87=E9=93=BE=E8=A1=A8=E3=80=91?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 高频面试系列/判断回文链表.md | 33 +++++++++++++++++++++++- 1 file changed, 32 insertions(+), 1 deletion(-) diff --git a/高频面试系列/判断回文链表.md b/高频面试系列/判断回文链表.md index 3e24068..4d0225a 100644 --- a/高频面试系列/判断回文链表.md +++ b/高频面试系列/判断回文链表.md @@ -237,4 +237,35 @@ p.next = reverse(q);

-======其他语言代码====== \ No newline at end of file +======其他语言代码====== + +C++版本： +```cpp + bool isPalindrome(ListNode* head) { + if (head == nullptr || head->next == nullptr) //为空或者只有一个节点时，直接判断为true + return true; + ListNode* slow = head, * fast = head; + while (fast != nullptr) {//首先找到中间节点 + slow = slow->next; + fast = fast->next == nullptr? fast->next:fast->next->next; //因为链表长度可能是奇数或偶数，所以需要进行判断 + } + + ListNode* temp = nullptr,* pre = nullptr;//pre始终保持后续链表的头部，temp节点则作为中间零时替换的节点 + while (slow != nullptr) {//利用头插法，将当前节点与后续链表断链处理，反转后半部分的链表 + temp = slow->next; + slow->next = pre;//建立连接 + pre = slow;//pre始终作为后续链表的头部 + slow = temp; + } + + while (head !=nullptr && pre != nullptr) {//同步进行比较 + if (head->val != pre->val) {//值有不一样的，说明不是回文联表，直接返回false了 + return false; + } + head = head->next;//head向下走，直到走到空 + pre = pre->next;//pre节点也向下走，直到走到空 + } + return true;//到此说明当前链表是回文链表返回true即可 + } + +``` From 6b55d3eceb7c3930552126d39e1da2d649493d9f Mon Sep 17 00:00:00 2001 From: Chenjie Xu Date: Wed, 11 Nov 2020 16:29:21 +0800 Subject: [PATCH 20/63] =?UTF-8?q?=E3=80=9072.=20=E7=BC=96=E8=BE=91?= =?UTF-8?q?=E8=B7=9D=E7=A6=BB=E3=80=91=E3=80=90Python=E3=80=91?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 动态规划系列/编辑距离.md | 28 ++++++++++++++++++++++++++++ 1 file changed, 28 insertions(+) diff --git a/动态规划系列/编辑距离.md b/动态规划系列/编辑距离.md index 27b8c22..ae6ff4e 100644 --- a/动态规划系列/编辑距离.md +++ b/动态规划系列/编辑距离.md @@ -290,3 +290,31 @@ class Node {

+ +[ChenjieXu](https://github.com/ChenjieXu) 提供Python版本代码： + +```python3 +def minDistance(word1, word2): + m, n = len(word1), len(word2) + # 创建 DP 数组 + dp = [[0] * (n + 1) for _ in range(m + 1)] + + # base case初始化 + for i in range(m + 1): + dp[i][0] = i + for j in range(n + 1): + dp[0][j] = j + + # 自底向上求解 + for i in range(1, m + 1): + for j in range(1, n + 1): + # 状态转移方程 + if word1[i - 1] == word2[j - 1]: + dp[i][j] = dp[i - 1][j - 1] + else: + dp[i][j] = min(dp[i - 1][j] + 1, + dp[i][j - 1] + 1, + dp[i - 1][j - 1] + 1) + # 储存着整个 word1 和 word2 的最小编辑距离 + return dp[m][n] +```` From 701622467245b7b960e57b3d4b1eea1b1b308e90 Mon Sep 17 00:00:00 2001 From: Ed Date: Wed, 11 Nov 2020 17:08:03 +0800 Subject: [PATCH 21/63] Add 452 python3 version --- .../贪心算法之区间调度问题.md | 43 ++++++++++++++++++- 1 file changed, 41 insertions(+), 2 deletions(-) diff --git a/动态规划系列/贪心算法之区间调度问题.md b/动态规划系列/贪心算法之区间调度问题.md index 70e27fc..8487083 100644 --- a/动态规划系列/贪心算法之区间调度问题.md +++ b/动态规划系列/贪心算法之区间调度问题.md @@ -1,4 +1,4 @@ -# 贪心算法之区间调度问题 +# 贪心算法之区间调度问题

@@ -158,4 +158,43 @@ int findMinArrowShots(int[][] intvs) {

-======其他语言代码====== \ No newline at end of file +======其他语言代码====== + +[Edwenc](https://github.com/Edwenc) 提供 第452题的python3 代码： + +```python3 + +class Solution: + def findMinArrowShots(self, points: List[List[int]]) -> int: + ### 思路是把气球的左右坐标看成是区间 + ### 用最少数量的箭 相当于 找不重叠区间的个数 + ### 因为重叠的区间 用一支箭就可以全部引爆气球 + + # 首先获得区间的个数 为0的话就不用移除 + n = len(points) + if n==0: + return 0 + + # 按照每个区间的右端点值进行排序 + sorted_point = sorted( points , key=lambda x: x[1] ) + + # 不重叠区间个数至少是1 + res = 1 + + # end是所有不重叠的区间中 最大的右端点 + # end的初始值即是sorted_list[0]的右端点 + end = sorted_point[0][1] + + # 从序号1开始往后找 + for i in range(1,n): + # start是当前区间左端点值 + start = sorted_point[i][0] + # 如果当前左端点比最大右端点都严格大了 + # 说明两区间严格不重叠 count+1 再更新end + if start > end: + res += 1 + end = sorted_point[i][1] + + return res + +``` \ No newline at end of file From 2ce19ed62bce8e79fc9f40b437ae33b3266fc311 Mon Sep 17 00:00:00 2001 From: Ed Date: Wed, 11 Nov 2020 17:17:08 +0800 Subject: [PATCH 22/63] Add 435 python3 version --- .../贪心算法之区间调度问题.md | 44 ++++++++++++++++++- 1 file changed, 42 insertions(+), 2 deletions(-) diff --git a/动态规划系列/贪心算法之区间调度问题.md b/动态规划系列/贪心算法之区间调度问题.md index 70e27fc..1865b27 100644 --- a/动态规划系列/贪心算法之区间调度问题.md +++ b/动态规划系列/贪心算法之区间调度问题.md @@ -1,4 +1,4 @@ -# 贪心算法之区间调度问题 +# 贪心算法之区间调度问题

@@ -158,4 +158,44 @@ int findMinArrowShots(int[][] intvs) {

-======其他语言代码====== \ No newline at end of file +======其他语言代码====== + +[Edwenc](https://github.com/Edwenc) 提供 第435题的python3 代码： + +```python3 + +class Solution: + def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int: + ### 思路是首先找到不重叠的区间的个数 + ### 然后再用总个数减去不重叠个数 + ### 获得的就是 需要移除的个数 + + # 首先获得区间的个数 为0的话就不用移除 + n = len(intervals) + if n==0: + return 0 + + # 按照每个区间的右端点值进行排序 + sorted_list = sorted( intervals , key=lambda x: x[1] ) + + # 不重叠区间个数至少是1 + count = 1 + + # end是所有不重叠的区间中 最大的右端点 + # end的初始值即是sorted_list[0]的右端点 + end = sorted_list[0][1] + + # 从1开始往后找 + for i in range(1,n): + # start是当前区间左端点值 + start = sorted_list[i][0] + # 如果左端点比右端点都大了（可能相等） + # 说明两区间不重叠 count+1 再更新end + if start>=end: + count += 1 + end = sorted_list[i][1] + + # 最后返回的是 需要移除的区间个数 + return n-count + +``` \ No newline at end of file From c09b441ba38c88adc75331087bd76bd4144a4a32 Mon Sep 17 00:00:00 2001 From: Ed Date: Wed, 11 Nov 2020 17:19:59 +0800 Subject: [PATCH 23/63] A little modification on comment --- 动态规划系列/贪心算法之区间调度问题.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/动态规划系列/贪心算法之区间调度问题.md b/动态规划系列/贪心算法之区间调度问题.md index 1865b27..05d9502 100644 --- a/动态规划系列/贪心算法之区间调度问题.md +++ b/动态规划系列/贪心算法之区间调度问题.md @@ -185,11 +185,11 @@ class Solution: # end的初始值即是sorted_list[0]的右端点 end = sorted_list[0][1] - # 从1开始往后找 + # 从1开始往后找 因为0在上面已经取过了 for i in range(1,n): # start是当前区间左端点值 start = sorted_list[i][0] - # 如果左端点比右端点都大了（可能相等） + # 如果当前左端点比最大右端点都大了（可能相等） # 说明两区间不重叠 count+1 再更新end if start>=end: count += 1 From b8441fd99e1aca9b2affa9143274d2b215f55a91 Mon Sep 17 00:00:00 2001 From: Zane Wang Date: Wed, 11 Nov 2020 20:28:34 +1100 Subject: [PATCH 24/63] =?UTF-8?q?=E4=B8=BA=2043-=E5=AD=97=E7=AC=A6?= =?UTF-8?q?=E4=B8=B2=E4=B9=98=E6=B3=95=20=E6=B7=BB=E5=8A=A0Java=E8=A7=A3?= =?UTF-8?q?=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 算法思维系列/字符串乘法.md | 45 ++++++++++++++++++++++++++- 1 file changed, 44 insertions(+), 1 deletion(-) diff --git a/算法思维系列/字符串乘法.md b/算法思维系列/字符串乘法.md index 0273744..6c7f67d 100644 --- a/算法思维系列/字符串乘法.md +++ b/算法思维系列/字符串乘法.md @@ -101,4 +101,47 @@ string multiply(string num1, string num2) {

-======其他语言代码====== \ No newline at end of file +======其他语言代码====== + +[Zane Wang](https://github.com/zanecat) 提供 Java 解法代码： +```java +public String multiply(String num1, String num2) { + // 初始化字符数组 + char[] s1 = num1.toCharArray(); + char[] s2 = num2.toCharArray(); + + // 结果长度最多为两字符串长度之和 + int[] res = new int[s1.length + s2.length]; + + // 从个位开始遍历，把两数字中每一位相乘 + for (int i = s1.length - 1; i >= 0; i--) { + for (int j = s2.length - 1; j >= 0; j--) { + // 计算乘积，并把乘积放在 res 对应的位置， 暂时不考虑进位 + res[i + j + 1] += (s1[i] - '0') * (s2[j] - '0'); + } + } + + // 从个位再次遍历，如果上一次遍历中两数乘积为两位数，进位并叠加到前面一位 + int carry = 0; + for (int i = res.length - 1; i >= 0; i--) { + int sum = res[i] + carry; + res[i] = sum % 10; + carry = sum / 10; + } + + //遍历res数组，构造最终答案字符串 + StringBuilder ans = new StringBuilder(); + int i = 0; + + // 首先找到不为0的第一位 + while (i < res.length - 1 && res[i] == 0) { + i++; + } + + // 将后面的数字附加到ans后面 + while (i < res.length) { + ans.append(res[i++]); + } + return ans.toString(); +} +``` \ No newline at end of file From 4e00179e784038b7f9aed3c54be38ee5e2b1b719 Mon Sep 17 00:00:00 2001 From: Ed Date: Wed, 11 Nov 2020 17:56:09 +0800 Subject: [PATCH 25/63] Add 450 Python3 version --- .../二叉搜索树操作集锦.md | 56 ++++++++++++++++++- 1 file changed, 54 insertions(+), 2 deletions(-) diff --git a/数据结构系列/二叉搜索树操作集锦.md b/数据结构系列/二叉搜索树操作集锦.md index 801b8fb..cac9c8c 100644 --- a/数据结构系列/二叉搜索树操作集锦.md +++ b/数据结构系列/二叉搜索树操作集锦.md @@ -1,4 +1,4 @@ -# 二叉搜索树操作集锦 +# 二叉搜索树操作集锦

@@ -310,4 +310,56 @@ void BST(TreeNode root, int target) {

-======其他语言代码====== \ No newline at end of file +======其他语言代码====== + +[Edwenc](https://github.com/Edwenc) 提供 leetcode第450题的python3 代码： + +```python3 + +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right + +class Solution: + def deleteNode(self, root: TreeNode, key: int) -> TreeNode: + # 如果没有树 直接返回None + if root == None: + return None + + # 如果要删除的结点 就是当前结点 + if root.val == key: + # 左子树为空 只有右子树需要被更新 直接返回 + if root.left == None: + return root.right + # 右子树为空 只有左子树需要被更新 直接返回 + if root.right== None: + return root.left + + # 找出此结点左子树的最大值 + # 用这个最大值 来代替当前结点 + # 再在左子树中递归地删除这个最大值结点 + big = self.getMax( root.left ) + root.val = big.val + root.left = self.deleteNode( root.left , big.val ) + + # 当前结点较大 它的左子树中需要删除节点 递归到左子树 + elif root.val > key: + root.left = self.deleteNode( root.left , key) + # 当前结点较小 它的右子树中需要删除节点 递归到右子树 + else: + root.right= self.deleteNode( root.right, key) + + return root + + # 辅助函数 + # 功能是找出此二叉搜索树中最大元素的结点 并返回此结点 + def getMax( self , node ): + # 一直找它的右子树 直到为空 + while node.right: + node = node.right + return node + +``` \ No newline at end of file From 9022e60cd2937dfdf714680b2a57a57c2376aae3 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=BD=97=E5=B7=8D=E8=80=80-LWW?= <694125111@qq.com> Date: Wed, 11 Nov 2020 19:59:07 +0800 Subject: [PATCH 26/63] =?UTF-8?q?Update=20=E7=83=A7=E9=A5=BC=E6=8E=92?= =?UTF-8?q?=E5=BA=8F.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 新增了Java解决方法 --- 算法思维系列/烧饼排序.md | 72 +++++++++++++++++++++++++++++- 1 file changed, 71 insertions(+), 1 deletion(-) diff --git a/算法思维系列/烧饼排序.md b/算法思维系列/烧饼排序.md index 513cbdc..cd163bf 100644 --- a/算法思维系列/烧饼排序.md +++ b/算法思维系列/烧饼排序.md @@ -149,4 +149,74 @@ void reverse(int[] arr, int i, int j) {

-======其他语言代码====== \ No newline at end of file +======其他语言代码====== + +[L-WEIWEI](https://github.com/L-WWEEII) 提供 第969题的 Java 代码： + +```java +class Solution { + public List pancakeSort(int[] A) { + List ans = new ArrayList(); + int len = A.length; + if(len == 0){ + return ans; + } + // maxIndex[0] == 当前轮次的最大元素， maxIndex[1] == 最大元素下标 + int[] maxIndex = new int[2]; + maxIndex[0] = Integer.MIN_VALUE; + int maxCount = 0; + // maxCount == len 时，说明完成了整个数组的最大值沉底操作， + while(maxCount < len - 1){ + maxCount = maxValueDown(A, maxIndex, maxCount, ans); + // 每做完一次最大值沉底操作，初始化最大元素值 + maxIndex[0] = Integer.MIN_VALUE; + } + return ans; + } + public int maxValueDown(int[] A, int[] maxIndex, int maxCount, List ans){ + // 遍历条件为 i < A.length - maxCount ， 每次最大值沉底时，maxCount + 1，因此下次遍历即可不对最后 maxCount 个元素做操作 + for(int i = 0; i < A.length - maxCount; i++){ + // 元素大于当前储存的元素时，将值与下标 copy 到 maxIndex 数组中 + if(A[i] > maxIndex[0]){ + maxIndex[0] = A[i]; + maxIndex[1] = i; + } + } + // 如果当前轮次最大元素的下标的下一位是上一轮次的最大下标，则不做翻转操作，直接返回 maxCount + 1 + if(maxIndex[1] + 1 == A.length - maxCount){ + return maxCount + 1; + } + // 使用最大值沉底时，当本轮最大值在首位时，不需要再将其先翻转至首位，所以不添加 + if(maxIndex[1] > 0){ + // 将该轮次要翻转的下标添加到结果集中，结果集中需要的是翻转的位置而不是下标，所以添加时下标得 + 1 + ans.add(maxIndex[1] + 1); + } + // 双指针原地交换数组中的值 + // 左指针指0 + int left = 0; + // 右指针指向当前轮次最大元素的下标 + int right = maxIndex[1]; + while(left < right){ + // 交换元素值 + A[left] += A[right]; + A[right] = A[left] - A[right]; + A[left] -= A[right]; + left++; + right--; + } + // 上面交换玩元素值后，当前轮次最大元素排在首位，再从上一轮次最大元素 - 1 的位置翻转 + // 则当前轮次的最大元素成功沉底 + ans.add(A.length - maxCount); + left = 0; + right = A.length - 1 - maxCount; + while(left < right){ + A[left] += A[right]; + A[right] = A[left] - A[right]; + A[left] -= A[right]; + left++; + right--; + } + return maxCount + 1; + } +} +``` From 58ce448d2b64b914373260e3fe0a5b19f69f37d6 Mon Sep 17 00:00:00 2001 From: Hoi Lung Lam Date: Wed, 11 Nov 2020 23:27:07 +1100 Subject: [PATCH 27/63] Added Python3 code for problem 3. --- 算法思维系列/滑动窗口技巧.md | 29 +++++++++++++++++++++++- 1 file changed, 28 insertions(+), 1 deletion(-) diff --git a/算法思维系列/滑动窗口技巧.md b/算法思维系列/滑动窗口技巧.md index e65a104..5d8768b 100644 --- a/算法思维系列/滑动窗口技巧.md +++ b/算法思维系列/滑动窗口技巧.md @@ -371,4 +371,31 @@ class Solution:

-======其他语言代码====== \ No newline at end of file +======其他语言代码====== + + + +第3题 Python3 代码（提供: [FaDrYL](https://github.com/FaDrYL) ）： +```Python3 +def lengthOfLongestSubstring(self, s: str) -> int: + # 子字符串 + sub = "" + largest = 0 + + # 循环字符串，将当前字符加入子字符串，并检查长度 + for i in range(len(s)): + if s[i] not in sub: + # 当前字符不存在于子字符串中，加入当前字符 + sub += s[i] + else: + # 如果当前子字符串的长度超过了之前的记录 + if len(sub) > largest: + largest = len(sub) + # 将子字符串从当前字符处+1切片至最后，并加入当前字符 + sub = sub[sub.find(s[i])+1:] + s[i] + + # 如果最后的子字符串长度超过了之前的记录 + if len(sub) > largest: + return len(sub) + return largest +``` From d1dcf4cc69efa50cfd78d18481ad41f7646680db Mon Sep 17 00:00:00 2001 From: userLF <1067605626@qq.com> Date: Wed, 11 Nov 2020 21:00:41 +0800 Subject: [PATCH 28/63] =?UTF-8?q?=E6=8F=90=E4=BA=A4=E4=BA=86=E5=85=A8?= =?UTF-8?q?=E6=8E=92=E5=88=97=E7=9A=84java=E4=BB=A3=E7=A0=81?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 高频面试系列/子集排列组合.md | 39 +++++++++++++++++++++++- 1 file changed, 38 insertions(+), 1 deletion(-) diff --git a/高频面试系列/子集排列组合.md b/高频面试系列/子集排列组合.md index 04fe3bb..bdb6bbe 100644 --- a/高频面试系列/子集排列组合.md +++ b/高频面试系列/子集排列组合.md @@ -283,5 +283,42 @@ void backtrack(int[] nums, LinkedList track) {

+======其他语言代码====== -======其他语言代码====== \ No newline at end of file +[userLF](https://github.com/userLF)提供全排列的java代码： + +```java +import java.util.ArrayList; +import java.util.List; + +class Solution { + List> res = new ArrayList<>(); + public List> permute(int[] nums) { + res.clear(); + dfs(nums, 0);// + return res; + } + + public void dfs(int[] n, int start) {//start表示要被替换元素的位置 + if( start >= n.length) { + List list = new ArrayList(); + for(int i : n) { + list.add(i); + } + res.add(list); + return; + } + + for(int i = start; i< n.length; i++) {//i从start开始，如果从start+1开始的话，会把当前序列遗漏掉直接保存了下一个序列 + int temp= n[i]; + n[i] = n[start]; + n[start] = temp; + dfs(n, start + 1);//递归下一个位置 + //回到上一个状态 + n[start] = n[i]; + n[i] = temp; + } + } +} + +``` \ No newline at end of file From 13afe04e8345d2fb7876c86f123561fb3ddb4b90 Mon Sep 17 00:00:00 2001 From: cchroot <843328872@qq.com> Date: Wed, 11 Nov 2020 21:21:20 +0800 Subject: [PATCH 29/63] =?UTF-8?q?=E3=80=90875.=20=E7=88=B1=E5=90=83?= =?UTF-8?q?=E9=A6=99=E8=95=89=E7=9A=84=E7=8F=82=E7=8F=82=E3=80=91=E3=80=90?= =?UTF-8?q?C++=E3=80=91?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 提供 C++ 解法 --- 高频面试系列/koko偷香蕉.md | 35 ++++++++++++++++++++++++++++- 1 file changed, 34 insertions(+), 1 deletion(-) diff --git a/高频面试系列/koko偷香蕉.md b/高频面试系列/koko偷香蕉.md index cc880dd..c44f083 100644 --- a/高频面试系列/koko偷香蕉.md +++ b/高频面试系列/koko偷香蕉.md @@ -169,4 +169,37 @@ for (int i = 0; i < n; i++)

-======其他语言代码====== \ No newline at end of file +======其他语言代码====== + +[cchroot](https://github.com/cchroot) 提供 C++ 代码： + +```c++ +class Solution { +public: + int minEatingSpeed(vector& piles, int H) { + // 二分法查找最小速度 + // 初始化最小速度为 1，最大速度为题目设定的最大值 10^9 + // 这里也可以遍历 piles 数组，获取数组中的最大值，设置 right 为数组中的最大值即可(因为每堆香蕉1小时吃完是最快的) + // log2(10^9) 约等于30，次数不多，所以这里暂时就不采取遍历获取最大值了 + int left = 1, right = pow(10, 9); + while (left < right) { // 二分法基本的防止溢出 + int mid = left + (right - left) / 2; + // 以 mid 的速度吃香蕉，是否能在 H 小时内吃完香蕉 + if (!canFinish(piles, mid, H)) + left = mid + 1; + else + right = mid; + } + return left; + } + + // 以 speed 的速度是否能把香蕉吃完 + bool canFinish(vector& piles, int speed, int H) { + int time = 0; + // 遍历累加时间 time + for (int p: piles) + time += (p - 1) / speed + 1; + return time <= H; // time 小于等于 H 说明能在 H 小时吃完返回 true, 否则返回 false + } +}; +``` From df9116647912d3a84ced39794a85384f100ba8bc Mon Sep 17 00:00:00 2001 From: LEO D PEN <33611404+LEODPEN@users.noreply.github.com> Date: Wed, 11 Nov 2020 21:25:36 +0800 Subject: [PATCH 30/63] =?UTF-8?q?leodpen=20Union-Find=E7=AE=97?= =?UTF-8?q?=E6=B3=95=E5=BA=94=E7=94=A8.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 算法思维系列/UnionFind算法应用.md | 78 ++++++++++++++++++++- 1 file changed, 77 insertions(+), 1 deletion(-) diff --git a/算法思维系列/UnionFind算法应用.md b/算法思维系列/UnionFind算法应用.md index fef484a..c1af151 100644 --- a/算法思维系列/UnionFind算法应用.md +++ b/算法思维系列/UnionFind算法应用.md @@ -20,6 +20,8 @@ [990.等式方程的可满足性](https://leetcode-cn.com/problems/surrounded-regions) +[261.以图判树](https://leetcode-cn.com/problems/graph-valid-tree/) + **-----------** 上篇文章很多读者对于 Union-Find 算法的应用表示很感兴趣，这篇文章就拿几道 LeetCode 题目来讲讲这个算法的巧妙用法。 @@ -250,4 +252,78 @@ boolean equationsPossible(String[] equations) {

-======其他语言代码====== \ No newline at end of file +======其他语言代码====== + +第261题的Java代码（提供：[LEODPEN](https://github.com/LEODPEN)） + +```java +class Solution { + + class DisjointSet { + + int count; // 连通分量的总个数 + int[] parent; // 每个节点的头节点（不一定是连通分量的最终头节点） + int[] size; // 每个连通分量的大小 + + public DisjointSet(int n) { + parent = new int[n]; + size = new int[n]; + // 初为n个连通分量，期望最后为1 + count = n; + for (int i = 0; i < n; i++) { + // 初始的连通分量只有该节点本身 + parent[i] = i; + size[i] = 1; + } + } + + /** + * @param first 节点1 + * @param second 节点2 + * @return 未连通 && 连通成功 + */ + public boolean union(int first, int second) { + // 分别找到包含first 和 second 的最终根节点 + int firstParent = findRootParent(first), secondParent = findRootParent(second); + // 相等说明已经处于一个连通分量，即说明有环 + if (firstParent == secondParent) return false; + // 将较小的连通分量融入较大的连通分量 + if (size[firstParent] >= size[secondParent]) { + parent[secondParent] = firstParent; + size[firstParent] += size[secondParent]; + } else { + parent[firstParent] = secondParent; + size[secondParent] += size[firstParent]; + } + // 连通分量已合并，count减少 + count--; + return true; + } + + /** + * @param node 某节点 + * @return 包含该节点的连通分量的最终根节点 + */ + private int findRootParent(int node) { + while (node != parent[node]) { + // 压缩路径 + parent[node] = parent[parent[node]]; + node = parent[node]; + } + return node; + } + } + + public boolean validTree(int n, int[][] edges) { + // 树的特性：节点数 = 边数 + 1 + if (edges.length != n - 1) return false; + graph_261.Solution.DisjointSet djs = new graph_261.Solution.DisjointSet(n); + for (int[] edg : edges) { + // 判断连通情况（如果合并的两个点在一个连通分量里，说明有环） + if (!djs.union(edg[0], edg[1])) return false; + } + // 是否全部节点均已相连 + return djs.count == 1; + } +} +``` From b946853ee8479241f349c721ac15c6922fc4879d Mon Sep 17 00:00:00 2001 From: cchroot <843328872@qq.com> Date: Wed, 11 Nov 2020 21:30:17 +0800 Subject: [PATCH 31/63] =?UTF-8?q?=E3=80=905.=20=E6=9C=80=E9=95=BF=E5=9B=9E?= =?UTF-8?q?=E6=96=87=E5=AD=90=E4=B8=B2=E3=80=91=E3=80=90Java=E3=80=91?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 提供 Java 代码 --- 高频面试系列/最长回文子串.md | 38 +++++++++++++++++++++++- 1 file changed, 37 insertions(+), 1 deletion(-) diff --git a/高频面试系列/最长回文子串.md b/高频面试系列/最长回文子串.md index 9f0f37f..d69b65e 100644 --- a/高频面试系列/最长回文子串.md +++ b/高频面试系列/最长回文子串.md @@ -131,4 +131,40 @@ string longestPalindrome(string s) {

-======其他语言代码====== \ No newline at end of file +======其他语言代码====== + +[cchromt](https://github.com/cchroot) 提供 Java 代码： + +```java +// 中心扩展算法 +class Solution { + public String longestPalindrome(String s) { + // 如果字符串长度小于2，则直接返回其本身 + if (s.length() < 2) { + return s; + } + String res = ""; + for (int i = 0; i < s.length() - 1; i++) { + // 以 s.charAt(i) 为中心的最长回文子串 + String s1 = palindrome(s, i, i); + // 以 s.charAt(i) 和 s.charAt(i+1) 为中心的最长回文子串 + String s2 = palindrome(s, i, i + 1); + res = res.length() > s1.length() ? res : s1; + res = res.length() > s2.length() ? res : s2; + } + return res; + } + + public String palindrome(String s, int left, int right) { + // 索引未越界的情况下，s.charAt(left) == s.charAt(right) 则继续向两边拓展 + while (left >= 0 && right < s.length() && s.charAt(left) == s.charAt(right)) { + left--; + right++; + } + // 这里要注意，跳出 while 循环时，恰好满足 s.charAt(i) != s.charAt(j)，因此截取的的字符串为[left+1, right-1] + return s.substring(left + 1, right); + } +} +``` + +做完这题，大家可以去看看 [647. 回文子串](https://leetcode-cn.com/problems/palindromic-substrings/) ,也是类似的题目 From 6f708d5b98f7d72411e0b0ec4c555c4f1584eefb Mon Sep 17 00:00:00 2001 From: ZSZ2018211261 <2955324023@qq.com> Date: Wed, 11 Nov 2020 22:35:43 +0800 Subject: [PATCH 32/63] =?UTF-8?q?=E5=A2=9E=E5=8A=A0=20=E3=80=9056.=20?= =?UTF-8?q?=E5=90=88=E5=B9=B6=E5=8C=BA=E9=97=B4=E3=80=91=E3=80=90java?= =?UTF-8?q?=E3=80=91=E9=A2=98=E8=A7=A3?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../区间调度问题之区间合并.md | 63 ++++++++++++++++++- 1 file changed, 62 insertions(+), 1 deletion(-) diff --git a/算法思维系列/区间调度问题之区间合并.md b/算法思维系列/区间调度问题之区间合并.md index e4ca2dc..7972184 100644 --- a/算法思维系列/区间调度问题之区间合并.md +++ b/算法思维系列/区间调度问题之区间合并.md @@ -90,5 +90,66 @@ def merge(intervals):

+======其他语言代码====== + +~~~java +class Solution { + /** + * 1. 先对区间集合进行排序（根据开始位置） + * 2. 合并的情况一共有三种 + * a. b. c. + * |---------| |--------| |--------| + * |---------| |--| |--------| + * a和b两种情况，合并取右边界大的值，c情况不合并 + * + */ + + private int[][] tmp; + + public int[][] merge(int[][] intervals) { + if(intervals == null ||intervals.length == 0)return new int[0][0]; + int length = intervals.length; + //将列表中的区间按照左端点升序排序 + // Arrays.sort(intervals,(v1,v2) -> v1[0]-v2[0]); + + this.tmp = new int[length][2]; + sort(intervals,0,length-1); + + int[][] ans = new int[length][2]; + int index = -1; + for(int[] interval:intervals){ + // 当结果数组是空是，或者当前区间的起始位置 > 结果数组中最后区间的终止位置（即上图情况c）； + // 则不合并，直接将当前区间加入结果数组。 + if(index == -1 || interval[0] > ans[index][1]){ + ans[++index] = interval; + }else{ + // 反之将当前区间合并至结果数组的最后区间（即上图情况a,b） + ans[index][1] = Math.max(ans[index][1],interval[1]); + } + } + return Arrays.copyOf(ans, index + 1); + } + + //归并排序 + public void sort(int[][] intervals,int l,int r){ + if(l >= r)return; + + int mid = l + (r-l)/2; + sort(intervals,l,mid); + sort(intervals,mid+1,r); + + //合并 + int i=l,j=mid+1; + for(int k=l;k<=r;k++){ + if(i>mid)tmp[k]=intervals[j++]; + else if(j>r)tmp[k]=intervals[i++]; + else if(intervals[i][0]>intervals[j][0])tmp[k] = intervals[j++]; + else tmp[k] = intervals[i++]; + } + + System.arraycopy(tmp,l,intervals,l,r-l+1); + } + +} +~~~ -======其他语言代码====== \ No newline at end of file From 97e5e1dee3df464f7deb181fba4139442d77ad26 Mon Sep 17 00:00:00 2001 From: hzs Date: Wed, 11 Nov 2020 17:44:49 +0800 Subject: [PATCH 33/63] =?UTF-8?q?=E3=80=90877.=E7=9F=B3=E5=AD=90=E6=B8=B8?= =?UTF-8?q?=E6=88=8F=E3=80=90Python=E3=80=91?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../动态规划之博弈问题.md | 55 ++++++++++++++++++- 1 file changed, 54 insertions(+), 1 deletion(-) diff --git a/动态规划系列/动态规划之博弈问题.md b/动态规划系列/动态规划之博弈问题.md index 9f52fa4..c85c100 100644 --- a/动态规划系列/动态规划之博弈问题.md +++ b/动态规划系列/动态规划之博弈问题.md @@ -215,4 +215,57 @@ int stoneGame(int[] piles) {

-======其他语言代码====== \ No newline at end of file +======其他语言代码====== + + + + + +python3版本 + +由[SCUHZS](https://github.com/brucecat)提供 + +这里采取的是三维的做法 + +```python +class Solution: + def stoneGame(self, piles: List[int]) -> bool: + n = len(piles) + + # 初始化一个n*n的矩阵 dp数组 + dp = [[None] * n for i in range(0, n)] + + # 在三角区域填充 + for i in range(n): + for j in range(i, n): + dp[i][j] = [0, 0] + + # 填入base case + for i in range(0, n): + dp[i][i][0] = piles[i] + dp[i][i][1] = 0 + + # 斜着遍历数组 + for l in range(2, n + 1): + for i in range(0, n-l+1): + j = l + i - 1 + + + # 先手选择最左边或最右边的分数 + left = piles[i] + dp[i + 1][j][1] + right = piles[j] + dp[i][j - 1][1] + + # 套用状态转移方程 + if left > right: + dp[i][j][0] = left + dp[i][j][1] = dp[i + 1][j][0] + else: + dp[i][j][0] = right + dp[i][j][1] = dp[i][j - 1][0] + + res = dp[0][n - 1] + + return res[0] - res[1] > 0 + +``` + From bc14b2e06201be75cf4a28b8d463e8e1f83f6608 Mon Sep 17 00:00:00 2001 From: hzs Date: Wed, 11 Nov 2020 17:50:05 +0800 Subject: [PATCH 34/63] =?UTF-8?q?=E3=80=90877.=E7=9F=B3=E5=AD=90=E6=B8=B8?= =?UTF-8?q?=E6=88=8F=E3=80=90Python=E3=80=91=E4=B8=80=E7=BB=B4dp?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../动态规划之博弈问题.md | 18 ++++++++++++++++++ 1 file changed, 18 insertions(+) diff --git a/动态规划系列/动态规划之博弈问题.md b/动态规划系列/动态规划之博弈问题.md index c85c100..90615ba 100644 --- a/动态规划系列/动态规划之博弈问题.md +++ b/动态规划系列/动态规划之博弈问题.md @@ -269,3 +269,21 @@ class Solution: ``` + + +压缩成一维数组，以减小空间复杂度，做法如下。 + +```python +class Solution: + def stoneGame(self, piles: List[int]) -> bool: + dp = piles.copy() + + for i in range(len(piles) - 1, -1, -1): # 从下往上遍历 + for j in range(i, len(piles)): # 从前往后遍历 + dp[j] = max(piles[i] - dp[j], piles[j] - dp[j - 1]) # 计算之后覆盖一维数组的对应位置 + + return dp[len(piles) - 1] > 0 + + +``` + From 36f59b6fe9ee832a7896e405c8bb76ae62b31e99 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=BD=97=E5=B7=8D=E8=80=80-LWW?= <694125111@qq.com> Date: Wed, 11 Nov 2020 22:27:58 +0800 Subject: [PATCH 35/63] =?UTF-8?q?Update=20=E6=89=93=E5=8D=B0=E7=B4=A0?= =?UTF-8?q?=E6=95=B0.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 增加了C++版本的代码 --- 高频面试系列/打印素数.md | 39 +++++++++++++++++++++++++++++- 1 file changed, 38 insertions(+), 1 deletion(-) diff --git a/高频面试系列/打印素数.md b/高频面试系列/打印素数.md index 2ab9e10..3a58074 100644 --- a/高频面试系列/打印素数.md +++ b/高频面试系列/打印素数.md @@ -178,4 +178,41 @@ int countPrimes(int n) {

-======其他语言代码====== \ No newline at end of file +======其他语言代码====== + +C++解法： +采用的算法是埃拉托斯特尼筛法 +埃拉托斯特尼筛法的具体内容就是：**要得到自然数n以内的全部素数，必须把不大于根号n的所有素数的倍数剔除，剩下的就是素数。** +同时考虑到大于2的偶数都不是素数，所以可以进一步优化成：**要得到自然数n以内的全部素数，必须把不大于根号n的所有素数的奇数倍剔除，剩下的奇数就是素数。** +此算法其实就是上面的Java解法所采用的。 + +这里提供C++的代码： +```C++ +class Solution { +public: + int countPrimes(int n) { + int res = 0; + bool prime[n+1]; + for(int i = 0; i < n; ++i) + prime[i] = true; + + for(int i = 2; i <= sqrt(n); ++i) //计数过程 + { //外循环优化，因为判断一个数是否为质数只需要整除到sqrt(n)，反推亦然 + if(prime[i]) + { + for(int j = i * i; j < n; j += i) //内循环优化，i*i之前的比如i*2，i*3等，在之前的循环中已经验证了 + { + prime[j] = false; + } + } + } + for (int i = 2; i < n; ++i) + if (prime[i]) res++; //最后遍历统计一遍，存入res + + return res; + } +}; +``` + + + From 6860be44b844b4147d68bb5501af87a59d390061 Mon Sep 17 00:00:00 2001 From: "shuhaofeng2@creditease.cn" Date: Wed, 11 Nov 2020 23:21:14 +0800 Subject: [PATCH 36/63] =?UTF-8?q?feat:=20=E4=B8=BA=2043-=E5=AD=97=E7=AC=A6?= =?UTF-8?q?=E4=B8=B2=E4=B9=98=E6=B3=95=20=E6=B7=BB=E5=8A=A0python=E8=A7=A3?= =?UTF-8?q?=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit feat: 为 43-字符串乘法 添加python解法1 feat: 为 43-字符串乘法 添加python解法3 --- 算法思维系列/字符串乘法.md | 31 ++++++++++++++++++++++++++- 1 file changed, 30 insertions(+), 1 deletion(-) diff --git a/算法思维系列/字符串乘法.md b/算法思维系列/字符串乘法.md index 0273744..8349f7a 100644 --- a/算法思维系列/字符串乘法.md +++ b/算法思维系列/字符串乘法.md @@ -101,4 +101,33 @@ string multiply(string num1, string num2) {

-======其他语言代码====== \ No newline at end of file +======其他语言代码====== + +[fengshuu](https://github.com/fengshuu) 提供 Python 解法代码： +```python +def multiply(num1: str, num2: str) -> str: + m, n = len(num1), len(num2) + # 结果最多为 m + n 位数 + res = [0] * (m + n) + # 从个位数开始逐位相乘 + for i in range(m-1, -1, -1): + for j in range(n-1, -1, -1): + mul = int(num1[i]) * int(num2[j]) + # 乘积在 res 对应的索引位置 + p1 = i + j + p2 = i + j + 1 + # 叠加到 res 上 + digit_sum = mul + res[p2] + res[p2] = digit_sum % 10 + res[p1] += digit_sum // 10 + + # 结果前缀可能存的 0（未使用的位） + i = 0 + while i < len(res) and res[i] == 0: + i += 1 + + # 将计算结果转化成字符串 + result_str = "".join(str(x) for x in res[i:]) + + return "0" if len(result_str) == 0 else result_str +``` \ No newline at end of file From edb2084f07cc23a9df50655da1c93dd690930dff Mon Sep 17 00:00:00 2001 From: zak Date: Wed, 11 Nov 2020 23:55:46 +0800 Subject: [PATCH 37/63] =?UTF-8?q?=E3=80=90496.=E4=B8=8B=E4=B8=80=E4=B8=AA?= =?UTF-8?q?=E6=9B=B4=E5=A4=A7=E5=85=83=E7=B4=A0I=E3=80=91?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 数据结构系列/单调栈.md | 56 ++++++++++++++++++++++++++++++++- 1 file changed, 55 insertions(+), 1 deletion(-) diff --git a/数据结构系列/单调栈.md b/数据结构系列/单调栈.md index fca0ca8..961e810 100644 --- a/数据结构系列/单调栈.md +++ b/数据结构系列/单调栈.md @@ -181,4 +181,58 @@ vector nextGreaterElements(vector& nums) {

-======其他语言代码====== \ No newline at end of file +======其他语言代码====== + +[ZakAnun](https://github.com/ZakAnun) 提供代码 + +```java +// 496.下一个更大元素 +// 暴力解法 +public int[] nextGreaterElement(int[] nums1, int[] nums2) { + int[] result = new int[nums1.length]; + for (int i = 0; i < nums1.length; i++) { + // 需要记录第一个数组每个元素在第二个数组中出现的位置 + int index = 0; + for (int j = 0; j < nums2.length; j++) { + if (nums1[i] == nums2[j]) { + index = j; + break; + } + } + // 根据找到的位置往后遍历，若符合条件则记录到结果数组 + for (int k = index; k < nums2.length; k++) { + if (nums2[k] > nums1[i]) { + result[i] = nums2[k]; + break; + } + } + // 判断若对应位置结果依然为默认值，则将其修改为 -1 + if (result[i] == 0) { + result[i] = -1; + } + } + return result; +} + +// 分析: 暴力解法中需要确定数组1中每个元素在数组2中的下标而需要进行额外的遍历导致时间复杂度升高， +// 但若能够先罗列出全部的结果，然后从结果集中获取数组1中每个元素对应的下一个更大元素，就可以节省这部分时间（这里需要引用 HashMap 帮助我们记录结果，以便根据数组1获取。 +// 单调栈解法 +public int[] nextGreaterElement(int[] nums1, int[] nums2) { + Stack stack = new Stack <>(); + HashMap map = new HashMap <>(); + int[] result = new int[nums1.length]; + for (int value : nums2) { + while (!stack.empty() && value > stack.peek()) { + map.put(stack.pop(), value); + } + stack.push(value); + } + while (!stack.empty()) { + map.put(stack.pop(), -1); + } + for (int i = 0; i < nums1.length; i++) { + result[i] = map.get(nums1[i]); + } + return result; +} +``` \ No newline at end of file From 9519492a8b9b9971a301be53dd006af09ba0689c Mon Sep 17 00:00:00 2001 From: "shuhaofeng2@creditease.cn" Date: Thu, 12 Nov 2020 00:45:38 +0800 Subject: [PATCH 38/63] =?UTF-8?q?feat:=20=E3=80=90969.=20=E7=85=8E?= =?UTF-8?q?=E9=A5=BC=E6=8E=92=E5=BA=8F=E3=80=91=E3=80=90C++=E3=80=91?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 算法思维系列/烧饼排序.md | 45 +++++++++++++++++++++++++++++- 1 file changed, 44 insertions(+), 1 deletion(-) diff --git a/算法思维系列/烧饼排序.md b/算法思维系列/烧饼排序.md index 513cbdc..5b9bf5d 100644 --- a/算法思维系列/烧饼排序.md +++ b/算法思维系列/烧饼排序.md @@ -149,4 +149,47 @@ void reverse(int[] arr, int i, int j) {

-======其他语言代码====== \ No newline at end of file +======其他语言代码====== + +[fengshuu](https://github.com/fengshuu) 提供 C++ 解法代码： +```cpp +class Solution { +public: + vector pancakeSort(vector& arr) { + sort(arr, arr.size()); + return res; + } + +private: + // 记录反转操作序列 + vector res; + void sort(vector& cakes, int n){ + // base case + if(n == 1) return; + + // 寻找最大饼的索引 + int maxCakeIndex = max_element(cakes.begin(), cakes.begin() + n) - cakes.begin(); + + // 下面进行把最大的饼放到最后的两次翻转 + // 如果最后一个饼就是最大的, 就不需要翻转, 直接进行下次递归 + if (maxCakeIndex == n-1){ + sort(cakes, n - 1); + return; + } + + // 第一次翻转, 将最大饼翻到最上面 + // 如果第一个饼本来就是最大的, 就不需要第一次翻转. + if (maxCakeIndex != 0) { + reverse(cakes.begin(), cakes.begin() + maxCakeIndex + 1); + res.push_back(maxCakeIndex + 1); + } + + // 第二次翻转，将最大饼翻到最下面 + reverse(cakes.begin(), cakes.begin() + n); + res.push_back(n); + + // 递归调用 + sort(cakes, n - 1); + } +}; +``` \ No newline at end of file From aa25e0d36c303bf49a93e27a2401f457edd6a4d8 Mon Sep 17 00:00:00 2001 From: "shuhaofeng2@creditease.cn" Date: Thu, 12 Nov 2020 00:55:23 +0800 Subject: [PATCH 39/63] =?UTF-8?q?feat:=20=E3=80=90969.=20=E7=85=8E?= =?UTF-8?q?=E9=A5=BC=E6=8E=92=E5=BA=8F=E3=80=91=E3=80=90C++=E3=80=91?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 算法思维系列/烧饼排序.md | 62 +++++++++++++++--------------- 1 file changed, 31 insertions(+), 31 deletions(-) diff --git a/算法思维系列/烧饼排序.md b/算法思维系列/烧饼排序.md index 5b9bf5d..c9e191d 100644 --- a/算法思维系列/烧饼排序.md +++ b/算法思维系列/烧饼排序.md @@ -157,39 +157,39 @@ class Solution { public: vector pancakeSort(vector& arr) { sort(arr, arr.size()); - return res; + return res; } - + private: - // 记录反转操作序列 + // 记录反转操作序列 vector res; - void sort(vector& cakes, int n){ - // base case - if(n == 1) return; - - // 寻找最大饼的索引 - int maxCakeIndex = max_element(cakes.begin(), cakes.begin() + n) - cakes.begin(); - - // 下面进行把最大的饼放到最后的两次翻转 - // 如果最后一个饼就是最大的, 就不需要翻转, 直接进行下次递归 - if (maxCakeIndex == n-1){ - sort(cakes, n - 1); - return; - } - - // 第一次翻转, 将最大饼翻到最上面 - // 如果第一个饼本来就是最大的, 就不需要第一次翻转. - if (maxCakeIndex != 0) { - reverse(cakes.begin(), cakes.begin() + maxCakeIndex + 1); - res.push_back(maxCakeIndex + 1); - } - - // 第二次翻转，将最大饼翻到最下面 - reverse(cakes.begin(), cakes.begin() + n); - res.push_back(n); - - // 递归调用 - sort(cakes, n - 1); - } + void sort(vector& cakes, int n){ + // base case + if(n == 1) return; + + // 寻找最大饼的索引 + int maxCakeIndex = max_element(cakes.begin(), cakes.begin() + n) - cakes.begin(); + + // 下面进行把最大的饼放到最后的两次翻转 + // 如果最后一个饼就是最大的, 就不需要翻转, 直接进行下次递归 + if (maxCakeIndex == n-1){ + sort(cakes, n - 1); + return; + } + + // 第一次翻转, 将最大饼翻到最上面 + // 如果第一个饼本来就是最大的, 就不需要第一次翻转. + if (maxCakeIndex != 0) { + reverse(cakes.begin(), cakes.begin() + maxCakeIndex + 1); + res.push_back(maxCakeIndex + 1); + } + + // 第二次翻转，将最大饼翻到最下面 + reverse(cakes.begin(), cakes.begin() + n); + res.push_back(n); + + // 递归调用 + sort(cakes, n - 1); + } }; ``` \ No newline at end of file From dd5749bb3bd66898ca6a06aec00da281e9dfd87f Mon Sep 17 00:00:00 2001 From: "shuhaofeng2@creditease.cn" Date: Thu, 12 Nov 2020 01:23:59 +0800 Subject: [PATCH 40/63] =?UTF-8?q?feat:=20=E3=80=90969.=20=E7=85=8E?= =?UTF-8?q?=E9=A5=BC=E6=8E=92=E5=BA=8F=E3=80=91=E3=80=90Python3=E3=80=91?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 算法思维系列/烧饼排序.md | 39 +++++++++++++++++++++++++++++- 1 file changed, 38 insertions(+), 1 deletion(-) diff --git a/算法思维系列/烧饼排序.md b/算法思维系列/烧饼排序.md index 513cbdc..67604ae 100644 --- a/算法思维系列/烧饼排序.md +++ b/算法思维系列/烧饼排序.md @@ -149,4 +149,41 @@ void reverse(int[] arr, int i, int j) {

-======其他语言代码====== \ No newline at end of file +======其他语言代码====== + +[fengshuu](https://github.com/fengshuu) 提供 Python3 解法代码： +```python +class Solution: + # 记录反转操作序列 + def __init__(self): + self.res = [] + + def pancakeSort(self, arr: List[int]) -> List[int]: + + self.sort(arr, len(arr)) + return self.res + + def sort(self, cakes: List[int], n: int): + # base case + if 1 == n: + return + + # 寻找最大饼的索引 + max_cake_index = cakes[:n].index(n) + + # 下面进行把最大的饼放到最后的两次翻转 + # 如果最后一个饼就是最大的, 就不需要翻转, 直接进行下次递归 + if max_cake_index != n - 1: + # 第一次翻转, 将最大饼翻到最上面 + # 如果第一个饼本来就是最大的, 就不需要第一次翻转. + if max_cake_index != 0: + cakes[:max_cake_index + 1] = cakes[:max_cake_index + 1][::-1] + self.res.append(max_cake_index + 1) + + # 第二次翻转，将最大饼翻到最下面 + cakes[:n] = cakes[:n][::-1] + self.res.append(n) + + # 递归调用 + self.sort(cakes, n - 1) +``` \ No newline at end of file From 8a2f374f63d568d3e45d99a6c0c4466d399ea2d5 Mon Sep 17 00:00:00 2001 From: "ningwei.shi" Date: Thu, 12 Nov 2020 09:21:54 +0800 Subject: [PATCH 41/63] =?UTF-8?q?=E5=A2=9E=E5=8A=A0=E6=B3=A8=E9=87=8A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 动态规划系列/抢房子.md | 2 ++ 1 file changed, 2 insertions(+) diff --git a/动态规划系列/抢房子.md b/动态规划系列/抢房子.md index d2c3f0f..081c583 100644 --- a/动态规划系列/抢房子.md +++ b/动态规划系列/抢房子.md @@ -311,7 +311,9 @@ class Solution: left = dp(root.left) right = dp(root.right) + # 抢当前，则两个下家不抢 do = root.val + left[0] + right[0] + # 不抢当前，则下家随意 do_not = max(left) + max(right) return do_not, do From d5b5f57380cf6a3f047a29253e811a10e5dcb498 Mon Sep 17 00:00:00 2001 From: LEO D PEN Date: Thu, 12 Nov 2020 12:00:29 +0800 Subject: [PATCH 42/63] =?UTF-8?q?Update=20UnionFind=E7=AE=97=E6=B3=95?= =?UTF-8?q?=E5=BA=94=E7=94=A8.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 算法思维系列/UnionFind算法应用.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/算法思维系列/UnionFind算法应用.md b/算法思维系列/UnionFind算法应用.md index c1af151..c3ce82f 100644 --- a/算法思维系列/UnionFind算法应用.md +++ b/算法思维系列/UnionFind算法应用.md @@ -317,7 +317,7 @@ class Solution { public boolean validTree(int n, int[][] edges) { // 树的特性：节点数 = 边数 + 1 if (edges.length != n - 1) return false; - graph_261.Solution.DisjointSet djs = new graph_261.Solution.DisjointSet(n); + DisjointSet djs = new DisjointSet(n); for (int[] edg : edges) { // 判断连通情况（如果合并的两个点在一个连通分量里，说明有环） if (!djs.union(edg[0], edg[1])) return false; From 6294fbcb734ad5f79bfa10ef46610f2f7e19f4d1 Mon Sep 17 00:00:00 2001 From: labuladong Date: Thu, 12 Nov 2020 14:26:22 +0800 Subject: [PATCH 43/63] fixurl --- 动态规划系列/动态规划之博弈问题.md | 2 +- .../动态规划设计：最长递增子序列.md | 10 +++++----- 动态规划系列/团灭股票问题.md | 4 ++-- 动态规划系列/最优子结构.md | 2 +- 动态规划系列/高楼扔鸡蛋问题.md | 2 +- 算法思维系列/滑动窗口技巧.md | 4 ++-- 6 files changed, 12 insertions(+), 12 deletions(-) diff --git a/动态规划系列/动态规划之博弈问题.md b/动态规划系列/动态规划之博弈问题.md index 90615ba..44246ab 100644 --- a/动态规划系列/动态规划之博弈问题.md +++ b/动态规划系列/动态规划之博弈问题.md @@ -22,7 +22,7 @@ 上一篇文章 [几道智力题](https://labuladong.gitbook.io/algo) 中讨论到一个有趣的「石头游戏」，通过题目的限制条件，这个游戏是先手必胜的。但是智力题终究是智力题，真正的算法问题肯定不会是投机取巧能搞定的。所以，本文就借石头游戏来讲讲「假设两个人都足够聪明，最后谁会获胜」这一类问题该如何用动态规划算法解决。 -博弈类问题的套路都差不多，下文举例讲解，其核心思路是在二维 dp 的基础上使用元组分别存储两个人的博弈结果。掌握了这个技巧以后，别人再问你什么俩海盗分宝石，俩人拿硬币的问题，你就告诉别人：我懒得想，直接给你写个算法算一下得了。 +博弈类问题的套路都差不多，下文参考 [这个 YouTube 视频](https://www.youtube.com/watch?v=WxpIHvsu1RI) 的思路讲解，其核心思路是在二维 dp 的基础上使用元组分别存储两个人的博弈结果。掌握了这个技巧以后，别人再问你什么俩海盗分宝石，俩人拿硬币的问题，你就告诉别人：我懒得想，直接给你写个算法算一下得了。 我们「石头游戏」改的更具有一般性： diff --git a/动态规划系列/动态规划设计：最长递增子序列.md b/动态规划系列/动态规划设计：最长递增子序列.md index c905a60..4613f84 100644 --- a/动态规划系列/动态规划设计：最长递增子序列.md +++ b/动态规划系列/动态规划设计：最长递增子序列.md @@ -10,8 +10,8 @@ ![](../pictures/souyisou.png) 相关推荐： - * [动态规划设计：最大子数组](../动态规划系列/最大子数组.md) - * [一文学会递归解题](../投稿/一文学会递归解题.md) + * [动态规划设计：最大子数组](https://labuladong.gitbook.io/algo) + * [一文学会递归解题](https://labuladong.gitbook.io/algo) 读完本文，你不仅学会了算法套路，还可以顺便去 LeetCode 上拿下如下题目： @@ -19,7 +19,7 @@ **-----------** -也许有读者看了前文 [动态规划详解](../动态规划系列/动态规划详解进阶.md)，学会了动态规划的套路：找到了问题的「状态」，明确了 `dp` 数组/函数的含义，定义了 base case；但是不知道如何确定「选择」，也就是不到状态转移的关系，依然写不出动态规划解法，怎么办？ +也许有读者看了前文 [动态规划详解](https://labuladong.gitbook.io/algo)，学会了动态规划的套路：找到了问题的「状态」，明确了 `dp` 数组/函数的含义，定义了 base case；但是不知道如何确定「选择」，也就是不到状态转移的关系，依然写不出动态规划解法，怎么办？ 不要担心，动态规划的难点本来就在于寻找正确的状态转移方程，本文就借助经典的「最长递增子序列问题」来讲一讲设计动态规划的通用技巧：**数学归纳思想**。 @@ -43,7 +43,7 @@ **我们的定义是这样的：`dp[i]` 表示以 `nums[i]` 这个数结尾的最长递增子序列的长度。** -PS：为什么这样定义呢？这是解决子序列问题的一个套路，后文[动态规划之子序列问题解题模板](../动态规划系列/子序列问题模板.md) 总结了几种常见套路。你读完本章所有的动态规划问题，就会发现 `dp` 数组的定义方法也就那几种。 +PS：为什么这样定义呢？这是解决子序列问题的一个套路，后文[动态规划之子序列问题解题模板](https://labuladong.gitbook.io/algo) 总结了几种常见套路。你读完本章所有的动态规划问题，就会发现 `dp` 数组的定义方法也就那几种。 根据这个定义，我们就可以推出 base case：`dp[i]` 初始值为 1，因为以 `nums[i]` 结尾的最长递增子序列起码要包含它自己。 @@ -164,7 +164,7 @@ public int lengthOfLIS(int[] nums) { 我们只要把处理扑克牌的过程编程写出来即可。每次处理一张扑克牌不是要找一个合适的牌堆顶来放吗，牌堆顶的牌不是**有序**吗，这就能用到二分查找了：用二分查找来搜索当前牌应放置的位置。 -PS：旧文[二分查找算法详解](../算法思维系列/二分查找详解.md)详细介绍了二分查找的细节及变体，这里就完美应用上了，如果没读过强烈建议阅读。 +PS：旧文[二分查找算法详解](https://labuladong.gitbook.io/algo)详细介绍了二分查找的细节及变体，这里就完美应用上了，如果没读过强烈建议阅读。 ```java public int lengthOfLIS(int[] nums) { diff --git a/动态规划系列/团灭股票问题.md b/动态规划系列/团灭股票问题.md index be2b463..61f64a3 100644 --- a/动态规划系列/团灭股票问题.md +++ b/动态规划系列/团灭股票问题.md @@ -16,7 +16,7 @@ 读完本文，你不仅学会了算法套路，还可以顺便去 LeetCode 上拿下如下题目： -[买卖股票的最佳时机](https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock/solution/) +[买卖股票的最佳时机](https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock) [买卖股票的最佳时机 II](https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock-ii/) @@ -32,7 +32,7 @@ 很多读者抱怨 LeetCode 的股票系列问题奇技淫巧太多，如果面试真的遇到这类问题，基本不会想到那些巧妙的办法，怎么办？**所以本文拒绝奇技淫巧，而是稳扎稳打，只用一种通用方法解决所用问题，以不变应万变**。 -这篇文章用状态机的技巧来解决，可以全部提交通过。不要觉得这个名词高大上，文学词汇而已，实际上就是 DP table，看一眼就明白了。 +这篇文章参考 [英文版高赞题解](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/discuss/108870/Most-consistent-ways-of-dealing-with-the-series-of-stock-problems) 的思路，用状态机的技巧来解决，可以全部提交通过。不要觉得这个名词高大上，文学词汇而已，实际上就是 DP table，看一眼就明白了。 先随便抽出一道题，看看别人的解法： diff --git a/动态规划系列/最优子结构.md b/动态规划系列/最优子结构.md index 2dc21f2..caf68cf 100644 --- a/动态规划系列/最优子结构.md +++ b/动态规划系列/最优子结构.md @@ -54,7 +54,7 @@ return result; 当然，上面这个例子太简单了，不过请读者回顾一下，我们做动态规划问题，是不是一直在求各种最值，本质跟我们举的例子没啥区别，无非需要处理一下重叠子问题。 -前文不[同定义不同解法](../动态规划系列/动态规划之四键键盘.md) 和 [高楼扔鸡蛋进阶](../动态规划系列/高楼扔鸡蛋问题.md) 就展示了如何改造问题，不同的最优子结构，可能导致不同的解法和效率。 +前文不[同定义不同解法](https://labuladong.gitbook.io/algo) 和 [高楼扔鸡蛋进阶](https://labuladong.gitbook.io/algo) 就展示了如何改造问题，不同的最优子结构，可能导致不同的解法和效率。 再举个常见但也十分简单的例子，求一棵二叉树的最大值，不难吧（简单起见，假设节点中的值都是非负数）： diff --git a/动态规划系列/高楼扔鸡蛋问题.md b/动态规划系列/高楼扔鸡蛋问题.md index 70c63a0..68bd9b8 100644 --- a/动态规划系列/高楼扔鸡蛋问题.md +++ b/动态规划系列/高楼扔鸡蛋问题.md @@ -243,7 +243,7 @@ def superEggDrop(self, K: int, N: int) -> int: return dp(K, N) ``` -这里就不展开其他解法了，留在下一篇文章 [高楼扔鸡蛋进阶](../动态规划系列/高楼扔鸡蛋进阶.md) +这里就不展开其他解法了，留在下一篇文章 [高楼扔鸡蛋进阶](https://labuladong.gitbook.io/algo) 我觉得吧，我们这种解法就够了：找状态，做选择，足够清晰易懂，可流程化，可举一反三。掌握这套框架学有余力的话，再去考虑那些奇技淫巧也不迟。 diff --git a/算法思维系列/滑动窗口技巧.md b/算法思维系列/滑动窗口技巧.md index e65a104..67e6539 100644 --- a/算法思维系列/滑动窗口技巧.md +++ b/算法思维系列/滑动窗口技巧.md @@ -12,8 +12,8 @@ **最新消息：关注公众号参与活动，有机会成为 [70k star 算法仓库](https://github.com/labuladong/fucking-algorithm) 的贡献者，机不可失时不再来**！ 相关推荐： -* [东哥吃葡萄时竟然吃出一道算法题！](../高频面试系列/吃葡萄.md) -* [如何寻找缺失的元素](../高频面试系列/消失的元素.md) +* [东哥吃葡萄时竟然吃出一道算法题！](https://labuladong.gitbook.io/algo) +* [如何寻找缺失的元素](https://labuladong.gitbook.io/algo) 读完本文，你不仅学会了算法套路，还可以顺便去 LeetCode 上拿下如下题目： From 4591b795796a287b76cf340adc1e69d1c10066d2 Mon Sep 17 00:00:00 2001 From: HuangXiao Date: Wed, 11 Nov 2020 11:51:34 +0800 Subject: [PATCH 44/63] Add java code for LCS --- 动态规划系列/最长公共子序列.md | 27 ++++++++++++++++++++- 1 file changed, 26 insertions(+), 1 deletion(-) diff --git a/动态规划系列/最长公共子序列.md b/动态规划系列/最长公共子序列.md index bee8947..39b13dd 100644 --- a/动态规划系列/最长公共子序列.md +++ b/动态规划系列/最长公共子序列.md @@ -147,4 +147,29 @@ else:

-======其他语言代码====== \ No newline at end of file +======其他语言代码====== + +[Shawn](https://github.com/Shawn-Hx) 提供 Java 代码： + +```java +public int longestCommonSubsequence(String text1, String text2) { + // 字符串转为char数组以加快访问速度 + char[] str1 = text1.toCharArray(); + char[] str2 = text2.toCharArray(); + + int m = str1.length, n = str2.length; + // 构建dp table，初始值默认为0 + int[][] dp = new int[m + 1][n + 1]; + // 状态转移 + for (int i = 1; i <= m; i++) + for (int j = 1; j <= n; j++) + if (str1[i - 1] == str2[j - 1]) + // 找到LCS中的字符 + dp[i][j] = dp[i-1][j-1] + 1; + else + dp[i][j] = Math.max(dp[i-1][j], dp[i][j-1]); + + return dp[m][n]; +} +``` + From 4d4a55f28ed891e8c98db8bb593ae54acf6d33cf Mon Sep 17 00:00:00 2001 From: Kian Kwok Date: Wed, 11 Nov 2020 11:37:25 +0800 Subject: [PATCH 45/63] =?UTF-8?q?Update=20=E5=8A=A8=E6=80=81=E8=A7=84?= =?UTF-8?q?=E5=88=92=E8=AE=BE=E8=AE=A1=EF=BC=9A=E6=9C=80=E9=95=BF=E9=80=92?= =?UTF-8?q?=E5=A2=9E=E5=AD=90=E5=BA=8F=E5=88=97.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit create problem-300 c++ --- ...态规划设计：最长递增子序列.md | 43 ++++++++++++++++++- 1 file changed, 42 insertions(+), 1 deletion(-) diff --git a/动态规划系列/动态规划设计：最长递增子序列.md b/动态规划系列/动态规划设计：最长递增子序列.md index 4613f84..a2a0460 100644 --- a/动态规划系列/动态规划设计：最长递增子序列.md +++ b/动态规划系列/动态规划设计：最长递增子序列.md @@ -215,4 +215,45 @@ public int lengthOfLIS(int[] nums) {

-======其他语言代码====== \ No newline at end of file +======其他语言代码====== + +[Kian](https://github.com/KianKw/) 提供 C++ 代码 + +```c++ +class Solution { +public: + int lengthOfLIS(vector& nums) { + /* len 为牌的数量 */ + int len = nums.size(); + vector top(len, 0); + /* 牌堆数初始化为0 */ + int piles = 0; + for (int i = 0; i < len; i++) { + /* nums[i] 为要处理的扑克牌 */ + int poker = nums[i]; + + /***** 搜索左侧边界的二分查找 *****/ + int left = 0, right = piles; + while (left < right) { + int mid = left + (right - left) / 2; + if (top[mid] > poker) { + right = mid; + } else if (top[mid] < poker) { + left = mid + 1; + } else if (top[mid] == poker) { + right = mid; + } + } + /*********************************/ + + /* 没找到合适的牌堆，新建一堆 */ + if (left == piles) + piles++; + /* 把这张牌放到牌堆顶 */ + top[left] = poker; + } + /* 牌堆数就是 LIS 长度 */ + return piles; + } +}; +``` From 8473c84f8d9207882efed9b45fdb5b1ac6153564 Mon Sep 17 00:00:00 2001 From: Lu Zhao Date: Tue, 10 Nov 2020 20:00:07 -0800 Subject: [PATCH 46/63] add 20 java --- 高频面试系列/合法括号判定.md | 26 +++++++++++++++++++++++- 1 file changed, 25 insertions(+), 1 deletion(-) diff --git a/高频面试系列/合法括号判定.md b/高频面试系列/合法括号判定.md index 32239bd..3bded5a 100644 --- a/高频面试系列/合法括号判定.md +++ b/高频面试系列/合法括号判定.md @@ -112,4 +112,28 @@ char leftOf(char c) {

-======其他语言代码====== \ No newline at end of file +======其他语言代码====== + +```java +//基本思想：每次遇到左括号时都将相对应的右括号'）'，']'或'}'推入堆栈 +//如果在字符串中出现右括号，则需要检查堆栈是否为空，以及顶部元素是否与该右括号相同。如果不是，则该字符串无效。 +//最后，我们还需要检查堆栈是否为空 +public boolean isValid(String s) { + Deque stack = new ArrayDeque<>(); + for(char c : s.toCharArray()){ + //是左括号就将相对应的右括号入栈 + if(c=='(') { + stack.offerLast(')'); + }else if(c=='{'){ + stack.offerLast('}'); + }else if(c=='['){ + stack.offerLast(']'); + }else if(stack.isEmpty() || stack.pollLast()!=c){//出现右括号，检查堆栈是否为空，以及顶部元素是否与该右括号相同 + return false; + } + } + return stack.isEmpty(); +} + +``` + From 780a1c46612684861613925295117ea7f85307b8 Mon Sep 17 00:00:00 2001 From: Ryan Deng Date: Tue, 10 Nov 2020 22:58:31 -0500 Subject: [PATCH 47/63] =?UTF-8?q?=E3=80=90141.=E7=8E=AF=E5=BD=A2=E9=93=BE?= =?UTF-8?q?=E8=A1=A8=E3=80=91=E3=80=90Python=E3=80=91?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 算法思维系列/双指针技巧.md | 23 ++++++++++++++++++++++- 1 file changed, 22 insertions(+), 1 deletion(-) diff --git a/算法思维系列/双指针技巧.md b/算法思维系列/双指针技巧.md index cf5227d..f5edbd4 100644 --- a/算法思维系列/双指针技巧.md +++ b/算法思维系列/双指针技巧.md @@ -230,4 +230,25 @@ void reverse(int[] nums) {

-======其他语言代码====== \ No newline at end of file +======其他语言代码====== + +[ryandeng32](https://github.com/ryandeng32/) 提供 Python 代码 +```python +class Solution: + def hasCycle(self, head: ListNode) -> bool: + # 检查链表头是否为None，是的话则不可能为环形 + if head is None: + return False + # 快慢指针初始化 + slow = fast = head + # 若链表非环形则快指针终究会遇到None，然后退出循环 + while fast.next and fast.next.next: + # 更新快慢指针 + slow = slow.next + fast = fast.next.next + # 快指针追上慢指针则链表为环形 + if slow == fast: + return True + # 退出循环，则链表有结束，不可能为环形 + return False +``` From 513fac4f4b9a44a71159118be5df08fa59c56245 Mon Sep 17 00:00:00 2001 From: FrankHui <757583912@qq.com> Date: Wed, 11 Nov 2020 12:54:40 +0800 Subject: [PATCH 48/63] =?UTF-8?q?=E3=80=90146.=20LRU=E7=BC=93=E5=AD=98?= =?UTF-8?q?=E6=9C=BA=E5=88=B6=E3=80=91=E3=80=90=20python3=E3=80=91?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit python3版本的LRU缓存机制 --- 高频面试系列/LRU算法.md | 32 +++++++++++++++++++++++++++++++- 1 file changed, 31 insertions(+), 1 deletion(-) diff --git a/高频面试系列/LRU算法.md b/高频面试系列/LRU算法.md index b9ca311..04ebea8 100644 --- a/高频面试系列/LRU算法.md +++ b/高频面试系列/LRU算法.md @@ -346,4 +346,34 @@ class LRUCache {

-======其他语言代码====== \ No newline at end of file +======其他语言代码====== +```python3 +""" +所谓LRU缓存，根本的难点在于记录最久被使用的键值对，这就设计到排序的问题， +在python中，天生具备排序功能的字典就是OrderDict。 +注意到，记录最久未被使用的键值对的充要条件是将每一次put/get的键值对都定义为 +最近访问，那么最久未被使用的键值对自然就会排到最后。 +如果你深入python OrderDict的底层实现，就会知道它的本质是个双向链表+字典。 +它内置支持了 +1. move_to_end来重排链表顺序，它可以让我们将最近访问的键值对放到最后面 +2. popitem来弹出键值对，它既可以弹出最近的，也可以弹出最远的，弹出最远的就是我们要的操作。 +""" +from collections import OrderedDict +class LRUCache: + def __init__(self, capacity: int): + self.capacity = capacity # cache的容量 + self.visited = OrderedDict() # python内置的OrderDict具备排序的功能 + def get(self, key: int) -> int: + if key not in self.visited: + return -1 + self.visited.move_to_end(key) # 最近访问的放到链表最后，维护好顺序 + return self.visited[key] + def put(self, key: int, value: int) -> None: + if key not in self.visited and len(self.visited) == self.capacity: + # last=False时，按照FIFO顺序弹出键值对 + # 因为我们将最近访问的放到最后，所以最远访问的就是最前的，也就是最first的，故要用FIFO顺序 + self.visited.popitem(last=False) + self.visited[key]=value + self.visited.move_to_end(key) # 最近访问的放到链表最后，维护好顺序 + +``` From 372b92bf7cd5bbbb00c96717806bed7f254da655 Mon Sep 17 00:00:00 2001 From: dekunma <53892579+dekunma@users.noreply.github.com> Date: Wed, 11 Nov 2020 14:21:23 +0800 Subject: [PATCH 49/63] =?UTF-8?q?=E3=80=9098.=20=E9=AA=8C=E8=AF=81?= =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=90=9C=E7=B4=A2=E6=A0=91=E3=80=91=E3=80=90?= =?UTF-8?q?C++=E3=80=91?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../二叉搜索树操作集锦.md | 34 ++++++++++++++++++- 1 file changed, 33 insertions(+), 1 deletion(-) diff --git a/数据结构系列/二叉搜索树操作集锦.md b/数据结构系列/二叉搜索树操作集锦.md index 801b8fb..b30ceeb 100644 --- a/数据结构系列/二叉搜索树操作集锦.md +++ b/数据结构系列/二叉搜索树操作集锦.md @@ -310,4 +310,36 @@ void BST(TreeNode root, int target) {

-======其他语言代码====== \ No newline at end of file +======其他语言代码====== +[dekunma](https://www.linkedin.com/in/dekun-ma-036a9b198/)提供第98题C++代码： +```C++ +/** + * Definition for a binary tree node. + * struct TreeNode { + * int val; + * TreeNode *left; + * TreeNode *right; + * TreeNode(int x) : val(x), left(NULL), right(NULL) {} + * }; + */ +class Solution { +public: + bool isValidBST(TreeNode* root) { + // 用helper method求解 + return isValidBST(root, nullptr, nullptr); + } + + bool isValidBST(TreeNode* root, TreeNode* min, TreeNode* max) { + // base case, root为nullptr + if (!root) return true; + + // 不符合BST的条件 + if (min && root->val <= min->val) return false; + if (max && root->val >= max->val) return false; + + // 向左右子树分别递归求解 + return isValidBST(root->left, min, root) + && isValidBST(root->right, root, max); + } +}; +``` From 3360ffda0fd458460a9c4908a0d7c0c58f539373 Mon Sep 17 00:00:00 2001 From: dragon_li <49151509+1097452462@users.noreply.github.com> Date: Thu, 12 Nov 2020 20:13:10 +0800 Subject: [PATCH 50/63] =?UTF-8?q?Update=20=E7=BC=BA=E5=A4=B1=E5=92=8C?= =?UTF-8?q?=E9=87=8D=E5=A4=8D=E7=9A=84=E5=85=83=E7=B4=A0.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 高频面试系列/缺失和重复的元素.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/高频面试系列/缺失和重复的元素.md b/高频面试系列/缺失和重复的元素.md index 52c8d78..9e0b550 100644 --- a/高频面试系列/缺失和重复的元素.md +++ b/高频面试系列/缺失和重复的元素.md @@ -1,4 +1,4 @@ -# 如何寻找缺失和重复的元素 +# 如何寻找缺失和重复的元素

@@ -165,4 +165,4 @@ class Solution { return new int[]{dup, missing}; } } -``` \ No newline at end of file +``` From 1b2ec335ea4b1ae65ba4cb1e92ff2b9e318b1084 Mon Sep 17 00:00:00 2001 From: dragon_li <49151509+1097452462@users.noreply.github.com> Date: Thu, 12 Nov 2020 20:15:23 +0800 Subject: [PATCH 51/63] =?UTF-8?q?Update=20=E7=BC=BA=E5=A4=B1=E5=92=8C?= =?UTF-8?q?=E9=87=8D=E5=A4=8D=E7=9A=84=E5=85=83=E7=B4=A0.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 高频面试系列/缺失和重复的元素.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/高频面试系列/缺失和重复的元素.md b/高频面试系列/缺失和重复的元素.md index 9e0b550..eb21f4c 100644 --- a/高频面试系列/缺失和重复的元素.md +++ b/高频面试系列/缺失和重复的元素.md @@ -148,7 +148,7 @@ class Solution { int n = nums.length; int dup = -1; for (int i = 0; i < n; i++) { - // 元素是从 1 开始的 + // 元素是从 1 开始的 int index = Math.abs(nums[i]) - 1; // nums[index] 小于 0 则说明重复访问 if (nums[index] < 0) @@ -160,7 +160,7 @@ class Solution { for (int i = 0; i < n; i++) // nums[i] 大于 0 则说明没有访问 if (nums[i] > 0) - // 将索引转换成元素 + // 将索引转换成元素 missing = i + 1; return new int[]{dup, missing}; } From 12fbf2910a9a99b5adcbe95f54787295e57d0bb3 Mon Sep 17 00:00:00 2001 From: ruiboliu Date: Tue, 10 Nov 2020 23:06:18 -0500 Subject: [PATCH 52/63] =?UTF-8?q?509=20=E6=96=90=E6=B3=A2=E9=82=A3?= =?UTF-8?q?=E5=A5=91=E6=95=B0=E5=88=97=20Python=203=20=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../动态规划详解进阶.md | 43 ++++++++++++++++++- 1 file changed, 42 insertions(+), 1 deletion(-) diff --git a/动态规划系列/动态规划详解进阶.md b/动态规划系列/动态规划详解进阶.md index bff29f4..8ec424b 100644 --- a/动态规划系列/动态规划详解进阶.md +++ b/动态规划系列/动态规划详解进阶.md @@ -366,4 +366,45 @@ PS：为啥 `dp` 数组初始化为 `amount + 1` 呢，因为凑成 `amount` 金

-======其他语言代码====== \ No newline at end of file +======其他语言代码====== + +[DapangLiu](https://github.com/DapangLiu) 提供 509. 斐波那契数 Python3 解法代码： + +递归写法 + +```python +class Solution: + def fib(self, N: int) -> int: + if N <= 1: + return N + return self.fib(N-1) + self.fib(N-2) +``` + +动态规划写法 + +```python +class Solution: + def fib(self, N: int) -> int: + if N == 0: + return 0 + # init + result = [0 for i in range(N+1)] + result[1] = 1 + + # status transition + for j in range(2, N+1): + result[j] = result[j-1] + result[j-2] + return result[-1] +``` + +动态规划写法 (状态压缩) + +```python +class Solution: + def fib(self, n: int) -> int: + # current status only depends on two previous status + dp_0, dp_1 = 0, 1 + for _ in range(n): + dp_0, dp_1 = dp_1, dp_0 + dp_1 + return dp_0 +``` \ No newline at end of file From 0ad9a50f645ff6328ea89fdd668bd1f8263dca2b Mon Sep 17 00:00:00 2001 From: Edwenc <45341488+Edwenc@users.noreply.github.com> Date: Tue, 17 Nov 2020 17:35:19 +0800 Subject: [PATCH 53/63] =?UTF-8?q?=E3=80=901143.=E6=9C=80=E9=95=BF=E5=85=AC?= =?UTF-8?q?=E5=85=B1=E5=AD=90=E5=BA=8F=E5=88=97=E3=80=91=E3=80=90C++=20?= =?UTF-8?q?=E3=80=91=20(#500)?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit * add 1143 C++ version * add comment Co-authored-by: Ed Co-authored-by: labuladong --- 动态规划系列/最长公共子序列.md | 37 ++++++++++++++++++++- 1 file changed, 36 insertions(+), 1 deletion(-) diff --git a/动态规划系列/最长公共子序列.md b/动态规划系列/最长公共子序列.md index 39b13dd..bff1721 100644 --- a/动态规划系列/最长公共子序列.md +++ b/动态规划系列/最长公共子序列.md @@ -1,4 +1,4 @@ -# 最长公共子序列 +# 最长公共子序列

@@ -149,6 +149,40 @@ else: ======其他语言代码====== + +[Edwenc](https://github.com/Edwenc) 提供 C++ 代码： + +```C++ +class Solution { +public: + int longestCommonSubsequence(string text1, string text2) { + // 先计算两条字符串的长度 + int m = text1.size(); + int n = text2.size(); + + // 构建dp矩阵 默认初始值0 + // 这里会多扩建一边和一列 + // 因为dp[i][j]的含义是：对于 s1[1..i] 和 s2[1..j]，它们的LCS长度是 dp[i][j]。 + // 所以当i或者j为零时 LCS的长度默认为0 + vector< vector > dp ( m+1 , vector ( n+1 , 0 ) ); + + // 状态转移 + // i、j都从1开始遍历 因为下面的操作中都会-1 相当于从0开始 + for ( int i=1 ; i Date: Tue, 10 Nov 2020 23:35:50 -0500 Subject: [PATCH 54/63] =?UTF-8?q?=E3=80=9042.=20=E6=8E=A5=E9=9B=A8?= =?UTF-8?q?=E6=B0=B4=E3=80=91=E3=80=90Java=E3=80=91?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 高频面试系列/接雨水.md | 97 ++++++++++++++++++++++++++++++++- 1 file changed, 96 insertions(+), 1 deletion(-) diff --git a/高频面试系列/接雨水.md b/高频面试系列/接雨水.md index e1fd625..a500aac 100644 --- a/高频面试系列/接雨水.md +++ b/高频面试系列/接雨水.md @@ -211,4 +211,99 @@ if (l_max < r_max) {

-======其他语言代码====== \ No newline at end of file +======其他语言代码====== + +[Yifan Zhang](https://github.com/FanFan0919) 提供 java 代码 + +**双指针解法**：时间复杂度 O(N)，空间复杂度 O(1) + +对cpp版本的解法有非常微小的优化。 +因为我们每次循环只会选 left 或者 right 处的柱子来计算，因此我们并不需要在每次循环中同时更新`maxLeft`和`maxRight`。 +我们可以先比较 `maxLeft` 和 `maxRight`，决定这次选择计算的柱子是 `height[left]` 或者 `height[right]` 后再更新对应的 `maxLeft` 或 `maxRight`。 +当然这并不会在时间上带来什么优化，只是提供一种思路。 + +```java +class Solution { + public int trap(int[] height) { + if (height == null || height.length == 0) return 0; + int left = 0, right = height.length - 1; + int maxLeft = height[left], maxRight = height[right]; + int res = 0; + + while (left < right) { + // 比较 maxLeft 和 maxRight，决定这次计算 left 还是 right 处的柱子 + if (maxLeft < maxRight) { + left++; + maxLeft = Math.max(maxLeft, height[left]); // update maxLeft + res += maxLeft - height[left]; + } else { + right--; + maxRight = Math.max(maxRight, height[right]); // update maxRight + res += maxRight - height[right]; + } + } + + return res; + } +} +``` + +附上暴力解法以及备忘录解法的 java 代码 + +**暴力解法**：时间复杂度 O(N^2)，空间复杂度 O(1) +```java +class Solution { + public int trap(int[] height) { + if (height == null || height.length == 0) return 0; + int n = height.length; + int res = 0; + // 跳过最左边和最右边的柱子，从第二个柱子开始 + for (int i = 1; i < n - 1; i++) { + int maxLeft = 0, maxRight = 0; + // 找右边最高的柱子 + for (int j = i; j < n; j++) { + maxRight = Math.max(maxRight, height[j]); + } + // 找左边最高的柱子 + for (int j = i; j >= 0; j--) { + maxLeft = Math.max(maxLeft, height[j]); + } + // 如果自己就是最高的话， + // maxLeft == maxRight == height[i] + res += Math.min(maxLeft, maxRight) - height[i]; + } + return res; + } +} +``` + +**备忘录解法**：时间复杂度 O(N)，空间复杂度 O(N) +```java +class Solution { + public int trap(int[] height) { + if (height == null || height.length == 0) return 0; + int n = height.length; + int res = 0; + // 数组充当备忘录 + int[] maxLeft = new int[n]; + int[] maxRight = new int[n]; + // 初始化 base case + maxLeft[0] = height[0]; + maxRight[n - 1] = height[n - 1]; + + // 从左向右计算 maxLeft + for (int i = 1; i < n; i++) { + maxLeft[i] = Math.max(maxLeft[i - 1], height[i]); + } + // 从右向左计算 maxRight + for (int i = n - 2; i >= 0; i--) { + maxRight[i] = Math.max(maxRight[i + 1], height[i]); + } + // 计算答案 + for (int i = 1; i < n; i++) { + res += Math.min(maxLeft[i], maxRight[i]) - height[i]; + } + return res; + } +} +``` \ No newline at end of file From 424f210d1560b0e16c46b83d34db65cc5cfa3120 Mon Sep 17 00:00:00 2001 From: Skylar Liang <31715464+SkylarLJY@users.noreply.github.com> Date: Tue, 17 Nov 2020 17:37:35 +0800 Subject: [PATCH 55/63] 20 Valid Parentheses python3 (#502) Co-authored-by: labuladong --- 高频面试系列/合法括号判定.md | 23 +++++++++++++++++++++++ 1 file changed, 23 insertions(+) diff --git a/高频面试系列/合法括号判定.md b/高频面试系列/合法括号判定.md index 3bded5a..6516aa1 100644 --- a/高频面试系列/合法括号判定.md +++ b/高频面试系列/合法括号判定.md @@ -114,6 +114,28 @@ char leftOf(char c) { ======其他语言代码====== +### Python3 +```python +def isValid(self, s: str) -> bool: + left = [] + leftOf = { + ')':'(', + ']':'[', + '}':'{' + } + for c in s: + if c in '([{': + left.append(c) + elif left and leftOf[c]==left[-1]: # 右括号 + left不为空 + 和最近左括号能匹配 + left.pop() + else: # 右括号 + （left为空 / 和堆顶括号不匹配） + return False + + # left中所有左括号都被匹配则return True 反之False + return not left +``` + + ```java //基本思想：每次遇到左括号时都将相对应的右括号'）'，']'或'}'推入堆栈 //如果在字符串中出现右括号，则需要检查堆栈是否为空，以及顶部元素是否与该右括号相同。如果不是，则该字符串无效。 @@ -137,3 +159,4 @@ public boolean isValid(String s) { ``` + From 7b6cab8386033457eaabc41a35e160ef0035fdf7 Mon Sep 17 00:00:00 2001 From: SherlockGuo <38394051+SherlockGuo@users.noreply.github.com> Date: Tue, 17 Nov 2020 17:49:13 +0800 Subject: [PATCH 56/63] =?UTF-8?q?=E3=80=90300.=E6=9C=80=E9=95=BF=E4=B8=8A?= =?UTF-8?q?=E5=8D=87=E5=AD=90=E5=BA=8F=E5=88=97=E3=80=91=E3=80=90Python?= =?UTF-8?q?=E3=80=91=20(#507)?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit * add python * add python * add python Co-authored-by: LockyGuo Co-authored-by: labuladong --- ...态规划设计：最长递增子序列.md | 46 +++++++++++++++++++ 1 file changed, 46 insertions(+) diff --git a/动态规划系列/动态规划设计：最长递增子序列.md b/动态规划系列/动态规划设计：最长递增子序列.md index a2a0460..edef473 100644 --- a/动态规划系列/动态规划设计：最长递增子序列.md +++ b/动态规划系列/动态规划设计：最长递增子序列.md @@ -217,6 +217,51 @@ public int lengthOfLIS(int[] nums) { ======其他语言代码====== +```python 动态规划 +class Solution: + def lengthOfLIS(self, nums: List[int]) -> int: + n = len(nums) + f = [1] * (n) + + for i in range(n): + for j in range(i): + if nums[j] < nums[i]: + f[i] = max(f[i], f[j] + 1) + + res = 0 + for i in range(n): + res = max(res, f[i]) + return res +``` + +```python 二分查找 +class Solution: + def lengthOfLIS(self, nums: List[int]) -> int: + stack = [] + + def find_index(num): + l, r = 0, len(stack) + while l < r: + mid = l + r >> 1 + if stack[mid] >= num: + r = mid + else: + l = mid + 1 + + return r + + + for num in nums: + if not stack or num > stack[-1]: + stack.append(num) + else: + position = find_index(num) + stack[position] = num + + return len(stack) +``` + + [Kian](https://github.com/KianKw/) 提供 C++ 代码 ```c++ @@ -257,3 +302,4 @@ public: } }; ``` + From 85301014bb879a33327dfa6643466c0184dd14c7 Mon Sep 17 00:00:00 2001 From: TCeason <33082201+TCeason@users.noreply.github.com> Date: Tue, 17 Nov 2020 17:56:31 +0800 Subject: [PATCH 57/63] add a cpp hash map version for stoneGame (#599) --- .../动态规划之博弈问题.md | 43 ++++++++++++++++++- 1 file changed, 42 insertions(+), 1 deletion(-) diff --git a/动态规划系列/动态规划之博弈问题.md b/动态规划系列/动态规划之博弈问题.md index 44246ab..f9082d8 100644 --- a/动态规划系列/动态规划之博弈问题.md +++ b/动态规划系列/动态规划之博弈问题.md @@ -221,7 +221,7 @@ int stoneGame(int[] piles) { -python3版本 +* python3版本 由[SCUHZS](https://github.com/brucecat)提供 @@ -287,3 +287,44 @@ class Solution: ``` +* C++ 版本 + +由 [TCeason](https://github.com/TCeason) 提供 + +这里采用 hash map 来解决问题 + +```cpp +class Solution { +public: + unordered_map memo; + + int dfs(vector &piles, int index) { + // 从两边向中间获取 + // index 值为 1/2 piles.size() 时可以停止算法 + if (index == piles.size() / 2) + return 0; + + // 减少计算，快速返回已有结果 + if (memo.count(index)) + return memo[index]; + + // 防止第一次取最右时越界 + int n = piles.size() - 1; + + // 先手选择最左边或最右边后的分数 + int l = piles[index] + dfs(piles, index + 1); + int r = piles[n - index] + dfs(piles, index + 1); + + // 返回先手左或右边的最高分 + return memo[index] = max(l, r); + } + + bool stoneGame(vector& piles) { + // 最佳发挥时： + // 先手得分 * 2 > 总大小 则先手者胜利 + return dfs(piles, 0) * 2 > accumulate(begin(piles), end(piles), 0); + } +}; + +``` + From 560ea5d9b9ac6fa3b5d8f82697871464c8784fb0 Mon Sep 17 00:00:00 2001 From: labuladong Date: Tue, 17 Nov 2020 19:14:05 +0800 Subject: [PATCH 58/63] =?UTF-8?q?=E6=90=9C=E4=B8=80=E6=90=9C?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- README.md | 2 ++ 1 file changed, 2 insertions(+) diff --git a/README.md b/README.md index c969002..45995ef 100644 --- a/README.md +++ b/README.md @@ -14,6 +14,8 @@ English version repo and Gitbook is on [english branch](https://github.com/labul

+![](pictures/souyisou.png) +

From 0b2efdc76576cce2b03971eceed00cba79765d48 Mon Sep 17 00:00:00 2001 From: labuladong Date: Wed, 18 Nov 2020 11:19:47 +0800 Subject: [PATCH 59/63] fixbug --- 动态规划系列/动态规划详解进阶.md | 4 +++- 数据结构系列/单调栈.md | 6 +++--- 算法思维系列/二分查找详解.md | 4 ++-- 算法思维系列/双指针技巧.md | 7 ++++++- 4 files changed, 14 insertions(+), 7 deletions(-) diff --git a/动态规划系列/动态规划详解进阶.md b/动态规划系列/动态规划详解进阶.md index 8ec424b..f51608e 100644 --- a/动态规划系列/动态规划详解进阶.md +++ b/动态规划系列/动态规划详解进阶.md @@ -146,6 +146,8 @@ int helper(vector& memo, int n) { ```cpp int fib(int N) { + if (N == 0) return 0; + if (N == 1) return 1; vector dp(N + 1, 0); // base case dp[1] = dp[2] = 1; @@ -200,7 +202,7 @@ int coinChange(int[] coins, int amount); 比如说 `k = 3`，面值分别为 1，2，5，总金额 `amount = 11`。那么最少需要 3 枚硬币凑出，即 11 = 5 + 5 + 1。 -你认为计算机应该如何解决这个问题？显然，就是把所有肯能的凑硬币方法都穷举出来，然后找找看最少需要多少枚硬币。 +你认为计算机应该如何解决这个问题？显然，就是把所有可能的凑硬币方法都穷举出来，然后找找看最少需要多少枚硬币。 **1、暴力递归** diff --git a/数据结构系列/单调栈.md b/数据结构系列/单调栈.md index fca0ca8..5cbdc4e 100644 --- a/数据结构系列/单调栈.md +++ b/数据结构系列/单调栈.md @@ -11,7 +11,7 @@ ![](../pictures/souyisou.png) 相关推荐： - * [回溯算法解题套路框架](https://labuladong.gitbook.io/algo) +* [回溯算法解题套路框架](https://labuladong.gitbook.io/algo) * [动态规划解题套路框架](https://labuladong.gitbook.io/algo) 读完本文，你不仅学会了算法套路，还可以顺便去 LeetCode 上拿下如下题目： @@ -20,7 +20,7 @@ [503.下一个更大元素II](https://leetcode-cn.com/problems/next-greater-element-ii) -[1118.一月有多少天](https://leetcode-cn.com/problems/number-of-days-in-a-month) +[739.每日温度](https://leetcode-cn.com/problems/daily-temperatures/) **-----------** @@ -82,7 +82,7 @@ vector nextGreaterElement(vector& nums) { ### 问题变形 -单调栈的使用技巧差不多了，来一个简单的变形，力扣第 1118 题「一月有多少天」： +单调栈的使用技巧差不多了，来一个简单的变形，力扣第 739 题「每日温度」： 给你一个数组 `T`，这个数组存放的是近几天的天气气温，你返回一个等长的数组，计算：**对于每一天，你还要至少等多少天才能等到一个更暖和的气温；如果等不到那一天，填 0**。 diff --git a/算法思维系列/二分查找详解.md b/算法思维系列/二分查找详解.md index 502962f..a413541 100644 --- a/算法思维系列/二分查找详解.md +++ b/算法思维系列/二分查找详解.md @@ -65,7 +65,7 @@ int binarySearch(int[] nums, int target) { ### 一、寻找一个数（基本的二分搜索） -这个场景是最简单的，肯能也是大家最熟悉的，即搜索一个数，如果存在，返回其索引，否则返回 -1。 +这个场景是最简单的，可能也是大家最熟悉的，即搜索一个数，如果存在，返回其索引，否则返回 -1。 ```java int binarySearch(int[] nums, int target) { @@ -104,7 +104,7 @@ int binarySearch(int[] nums, int target) { `while(left <= right)` 的终止条件是 `left == right + 1`，写成区间的形式就是 `[right + 1, right]`，或者带个具体的数字进去 `[3, 2]`，可见**这时候区间为空**，因为没有数字既大于等于 3 又小于等于 2 的吧。所以这时候 while 循环终止是正确的，直接返回 -1 即可。 -`while(left < right)` 的终止条件是 `left == right`，写成区间的形式就是 `[left, right]`，或者带个具体的数字进去 `[2, 2]`，**这时候区间非空**，还有一个数 2，但此时 while 循环终止了。也就是说这区间 `[2, 2]` 被漏掉了，索引 2 没有被搜索，如果这时候直接返回 -1 就是错误的。 +`while(left < right)` 的终止条件是 `left == right`，写成区间的形式就是 `[right, right]`，或者带个具体的数字进去 `[2, 2]`，**这时候区间非空**，还有一个数 2，但此时 while 循环终止了。也就是说这区间 `[2, 2]` 被漏掉了，索引 2 没有被搜索，如果这时候直接返回 -1 就是错误的。 当然，如果你非要用 `while(left < right)` 也可以，我们已经知道了出错的原因，就打个补丁好了： diff --git a/算法思维系列/双指针技巧.md b/算法思维系列/双指针技巧.md index f5edbd4..ff2aebe 100644 --- a/算法思维系列/双指针技巧.md +++ b/算法思维系列/双指针技巧.md @@ -20,7 +20,7 @@ [141.环形链表II](https://leetcode-cn.com/problems/linked-list-cycle-ii) -[167.两数之和 II - 输入有序数组](https://leetcode-cn.com/problems/two-sum) +[167.两数之和 II - 输入有序数组](https://leetcode-cn.com/problems/two-sum-ii-input-array-is-sorted) **-----------** @@ -80,6 +80,11 @@ ListNode detectCycle(ListNode head) { if (fast == slow) break; } // 上面的代码类似 hasCycle 函数 + if (fast == null || fast.next == null) { + // fast 遇到空指针说明没有环 + return null; + } + slow = head; while (slow != fast) { fast = fast.next; From fb6ada33111aefc22de515579f3e680f5614b8fe Mon Sep 17 00:00:00 2001 From: labuladong Date: Thu, 3 Dec 2020 10:43:18 +0800 Subject: [PATCH 60/63] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=E5=87=BA=E7=89=88?= =?UTF-8?q?=E6=8E=A8=E5=B9=BF?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- README.md | 25 ++++++++++++++++++------- 出版推广1.jpeg | Bin 0 -> 122844 bytes 2 files changed, 18 insertions(+), 7 deletions(-) create mode 100644 出版推广1.jpeg diff --git a/README.md b/README.md index 45995ef..0858fff 100644 --- a/README.md +++ b/README.md @@ -16,6 +16,8 @@ English version repo and Gitbook is on [english branch](https://github.com/labul ![](pictures/souyisou.png) +好消息，《labuladong 的算法小抄》纸质书出版啦！关注公众号查看详情👆 +

@@ -27,27 +29,36 @@ English version repo and Gitbook is on [english branch](https://github.com/labul 只想要答案的话很容易，题目评论区五花八门的答案，动不动就秀 python 一行代码解决，有那么多人点赞。问题是，你去做算法题，是去学习编程语言的奇技淫巧的，还是学习算法思维的呢？你的快乐，到底源自复制别人的一行代码通过测试，已完成题目 +1，还是源自自己通过逻辑推理和算法框架不看答案写出解法？ -网上总有大佬喷我，说我写这玩意太基础了，根本没必要啰嗦。我只能说大家刷算法就是找工作吃饭的，不是打竞赛的，我也是一路摸爬滚打过来的，我们要的是清楚明白有所得，不是故弄玄虚无所指。不想办法做到通俗易懂，难道要上来先把《算法导论》吹上天，然后把人家都心怀敬仰地劝退？ +网上总有大佬喷我，说我写的东西太基础，要么说不能借助框架思维来学习算法。我只能说大家刷算法就是找工作吃饭的，不是打竞赛的，我也是一路摸爬滚打过来的，我们要的是清楚明白有所得，不是故弄玄虚无所指。 + +不想办法做到通俗易懂，难道要上来先把《算法导论》吹上天，然后把人家都心怀敬仰地劝退？ **做啥事情做多了，都能发现套路的，我把各种算法套路框架总结出来，相信可以帮助其他人少走弯路**。我这个纯靠自学的小童鞋，花了一年时间刷题和总结，自己写了一份算法小抄，后面有目录，这里就不废话了。 ### 使用方法 -1、**先给本仓库点个 star，满足一下我的虚荣心**，文章质量绝对值你一个 star。我还在继续创作，给我一点继续写文的动力，感谢。 +**1、先给本仓库点个 star，满足一下我的虚荣心**，文章质量绝对值你一个 star。我还在继续创作，给我一点继续写文的动力，感谢。 -2、**建议收藏我的 Gitbook 网站，每篇文章开头都有对应的力扣题目链接，可以边看文章边刷题**： +**2、建议收藏我的在线网站，每篇文章开头都有对应的力扣题目链接，可以边看文章边刷题**： -Gitbook 地址：https://labuladong.gitbook.io/algo/ +Gitbook 地址：https://labuladong.gitbook.io/algo -3、建议关注我的公众号 **labuladong**，坚持高质量原创，说是最良心最硬核的技术公众号都不为过。本仓库的文章就是从公众号里整理出来的**一部分**内容，公众号后台回复关键词【电子书】可以获得这份小抄的完整版本；回复【加群】可以加入我们的刷题群，和大家一起讨论算法问题，分享内推机会： +有部分读者反映 GitBook 访问速度较慢，我特意部署了两个镜像站点，大家可根据网络情况自行选择： + +GitHub Pages 地址：https://labuladong.github.io/algo + +Gitee Pages 地址：https://labuladong.gitee.io/algo + + +**3、建议关注我的公众号 labuladong，坚持高质量原创，说是最良心最硬核的技术公众号都不为过**。本仓库的文章就是从公众号里整理出来的**一部分**内容，公众号可以查看更多内容；公众号后台回复关键词【加群】可以加入我们的刷题群，和大家一起讨论算法问题，分享内推机会：

-4、欢迎关注 [我的知乎](https://www.zhihu.com/people/labuladong)。 +**4、欢迎关注 [我的知乎](https://www.zhihu.com/people/labuladong)**。 -我一直在写优质文章，但是后续的文章只发布到公众号/gitbook/知乎，不能开放到 GitHub。因为本仓库太火了，很多人直接拿我的文章去开付费专栏，价格还不便宜，我这免费写给您看，何必掏冤枉钱呢？所以多多关注本作者，多多宣传，谁也不希望劣币驱逐良币不是么？ +我一直在写优质文章，但是后续的文章只发布到公众号/网站/知乎，不能开放到 GitHub。因为本仓库太火了，很多人直接拿我的文章去开付费专栏，价格还不便宜，我这免费写给您看，何必掏冤枉钱呢？所以多多关注本作者，多多宣传，谁也不希望劣币驱逐良币不是么？ 其他的先不多说了，直接上干货吧，我们一起搞定 LeetCode，感受一下支配算法的乐趣。 diff --git a/出版推广1.jpeg b/出版推广1.jpeg new file mode 100644 index 0000000000000000000000000000000000000000..6205bfee59f2363a6117d26ac667b50d06cb43cf GIT binary patch literal 122844 zcmeFY2UJtvmp^&|5di@~iXspdkPaeMDZxS)>AgfnqzQ=hmVihRkO&GWC_(8pA|ObW zE&|d)Kw6Lvl29+Cz3a@ZneTk({b$YZ{oi|Qy|;dWla+gOa_-*y?6c1SbMr#LvdIR6YnLkIol2t6Y`JtHRzGYjWG{GxsV*pJbm z8BfsBoCj#xY3SH#sBHiY0BGnTYX1=YZ!a2Jh{mG~#~7KIp$=6i09qP4I@%+2f2f9f zhe6*1N7(5(&dFXqdeZm~!+CGcOA$%$jtO2XZ{{)?zzfOUee{@-iJRxtXX=QEWc+cs+vx}>nkFTG9KwwaCYC5BEv;?s9i3mhz6}lykBpA}7@wHK&Hq|hTv}dP 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b/README.md @@ -3,7 +3,7 @@ English version repo and Gitbook is on [english branch](https://github.com/labul # labuladong 的算法小抄

-Website +Website GitHub

@@ -43,7 +43,7 @@ English version repo and Gitbook is on [english branch](https://github.com/labul Gitbook 地址：https://labuladong.gitbook.io/algo -有部分读者反映 GitBook 访问速度较慢，我特意部署了两个镜像站点，大家可根据网络情况自行选择： +GitBook 在国内访问速度很慢，且常被攻击，我特意部署了两个镜像站点，大家可根据网络情况自行选择： GitHub Pages 地址：https://labuladong.github.io/algo From 6e2126a860659a7165b61135ae6c36b172774f87 Mon Sep 17 00:00:00 2001 From: BruceCat <41747005+brucecat@users.noreply.github.com> Date: Sun, 7 Mar 2021 17:00:22 +0800 Subject: [PATCH 62/63] =?UTF-8?q?=E3=80=90739.=20Daily=20Temperatures?= =?UTF-8?q?=E3=80=91=E3=80=90Java=E3=80=91?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 数据结构系列/单调栈.md | 14 +++++++++++--- 1 file changed, 11 insertions(+), 3 deletions(-) diff --git a/数据结构系列/单调栈.md b/数据结构系列/单调栈.md index feedf44..28b4f91 100644 --- a/数据结构系列/单调栈.md +++ b/数据结构系列/单调栈.md @@ -8,9 +8,11 @@

-![](../pictures/souyisou.png) + +![](pictures/souyisou.png) 相关推荐： + * [回溯算法解题套路框架](https://labuladong.gitbook.io/algo) * [动态规划解题套路框架](https://labuladong.gitbook.io/algo) @@ -50,7 +52,7 @@ vector nextGreaterElement(vector& nums); 这个问题可以这样抽象思考：把数组的元素想象成并列站立的人，元素大小想象成人的身高。这些人面对你站成一列，如何求元素「2」的 Next Greater Number 呢？很简单，如果能够看到元素「2」，那么他后面可见的第一个人就是「2」的 Next Greater Number，因为比「2」小的元素身高不够，都被「2」挡住了，第一个露出来的就是答案。 -![](../pictures/%E5%8D%95%E8%B0%83%E6%A0%88/1.jpeg) +![](pictures/单调栈/1.jpeg) 这个情景很好理解吧？带着这个抽象的情景，先来看下代码。 @@ -142,7 +144,7 @@ while (true) { **对于这种需求，常用套路就是将数组长度翻倍**： -![](../pictures/%E5%8D%95%E8%B0%83%E6%A0%88/2.jpeg) +![](pictures/单调栈/2.jpeg) 这样，元素 3 就可以找到元素 4 作为 Next Greater Number 了，而且其他的元素都可以被正确地计算。 @@ -182,6 +184,10 @@ vector nextGreaterElements(vector& nums) {

======其他语言代码====== + +### java + +```java // 739. Daily Temperatures class Solution { public int[] dailyTemperatures(int[] T) { @@ -199,3 +205,5 @@ class Solution { return ans; } } +``` + From 20ce356b32a443b0656346197cf6e6603696cf2e Mon Sep 17 00:00:00 2001 From: BruceCat <41747005+brucecat@users.noreply.github.com> Date: Sun, 7 Mar 2021 17:04:58 +0800 Subject: [PATCH 63/63] =?UTF-8?q?=E3=80=90739.=20Daily=20Temperatures?= =?UTF-8?q?=E3=80=91=E3=80=90Java=E3=80=91?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 【739. Daily Temperatures】【Java】 --- 数据结构系列/单调栈.md | 9 +++------ 1 file changed, 3 insertions(+), 6 deletions(-) diff --git a/数据结构系列/单调栈.md b/数据结构系列/单调栈.md index 28b4f91..d49168b 100644 --- a/数据结构系列/单调栈.md +++ b/数据结构系列/单调栈.md @@ -8,11 +8,9 @@

- -![](pictures/souyisou.png) +![](../pictures/souyisou.png) 相关推荐： - * [回溯算法解题套路框架](https://labuladong.gitbook.io/algo) * [动态规划解题套路框架](https://labuladong.gitbook.io/algo) @@ -52,7 +50,7 @@ vector nextGreaterElement(vector& nums); 这个问题可以这样抽象思考：把数组的元素想象成并列站立的人，元素大小想象成人的身高。这些人面对你站成一列，如何求元素「2」的 Next Greater Number 呢？很简单，如果能够看到元素「2」，那么他后面可见的第一个人就是「2」的 Next Greater Number，因为比「2」小的元素身高不够，都被「2」挡住了，第一个露出来的就是答案。 -![](pictures/单调栈/1.jpeg) +![](../pictures/%E5%8D%95%E8%B0%83%E6%A0%88/1.jpeg) 这个情景很好理解吧？带着这个抽象的情景，先来看下代码。 @@ -144,7 +142,7 @@ while (true) { **对于这种需求，常用套路就是将数组长度翻倍**： -![](pictures/单调栈/2.jpeg) +![](../pictures/%E5%8D%95%E8%B0%83%E6%A0%88/2.jpeg) 这样，元素 3 就可以找到元素 4 作为 Next Greater Number 了，而且其他的元素都可以被正确地计算。 @@ -184,7 +182,6 @@ vector nextGreaterElements(vector& nums) {

======其他语言代码====== - ### java ```java