From 178cf44b202bf10707fed8825d3bf23b8e9d8f8b Mon Sep 17 00:00:00 2001 From: "xeniaxie(xl)" Date: Sat, 26 Jun 2021 20:33:27 +0800 Subject: [PATCH 1/8] =?UTF-8?q?0501=E4=BA=8C=E5=8F=89=E6=90=9C=E7=B4=A2?= =?UTF-8?q?=E6=A0=91=E4=B8=AD=E7=9A=84=E4=BC=97=E6=95=B0JavaScript?= =?UTF-8?q?=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0501.二叉搜索树中的众数.md | 74 +++++++++++++++++++- 1 file changed, 73 insertions(+), 1 deletion(-) diff --git a/problems/0501.二叉搜索树中的众数.md b/problems/0501.二叉搜索树中的众数.md index 3ca7d892..74c53458 100644 --- a/problems/0501.二叉搜索树中的众数.md +++ b/problems/0501.二叉搜索树中的众数.md @@ -522,7 +522,79 @@ func traversal(root *TreeNode,result *[]int,pre *TreeNode){//遍历统计 } } ``` - +JavaScript版本： +使用额外空间map的方法： +```javascript +var findMode = function(root) { + // 使用递归中序遍历 + let map = new Map(); + // 1. 确定递归函数以及函数参数 + const traverTree = function(root) { + // 2. 确定递归终止条件 + if(root === null) { + return ; + } + traverTree(root.left); + // 3. 单层递归逻辑 + map.set(root.val,map.has(root.val)?map.get(root.val)+1:1); + traverTree(root.right); + } + traverTree(root); + //上面把数据都存储到map + //下面开始寻找map里面的 + // 定义一个最大出现次数的初始值为root.val的出现次数 + let maxCount = map.get(root.val); + // 定义一个存放结果的数组res + let res = []; + for(let [key,value] of map) { + // 如果当前值等于最大出现次数就直接在res增加该值 + if(value === maxCount) { + res.push(key); + } + // 如果value的值大于原本的maxCount就清空res的所有值，因为找到了更大的 + if(value>maxCount) { + res = []; + maxCount = value; + res.push(key); + } + } + return res; +}; +``` +不使用额外空间，利用二叉树性质，中序遍历(有序)： +```javascript +var findMode = function(root) { + // 不使用额外空间，使用中序遍历,设置出现最大次数初始值为1 + let count = 0,maxCount = 1; + let pre = root,res = []; + // 1.确定递归函数及函数参数 + const travelTree = function(cur) { + // 2. 确定递归终止条件 + if(cur === null) { + return ; + } + travelTree(cur.left); + // 3. 单层递归逻辑 + if(pre.val === cur.val) { + count++; + }else { + count = 1; + } + pre = cur; + if(count === maxCount) { + res.push(cur.val); + } + if(count > maxCount) { + res = []; + maxCount = count; + res.push(cur.val); + } + travelTree(cur.right); + } + travelTree(root); + return res; +}; +``` ----------------------- From f3b21c7984039eaa37c8da08eac725e6c8d1e11b Mon Sep 17 00:00:00 2001 From: "xeniaxie(xl)" Date: Sat, 26 Jun 2021 21:12:39 +0800 Subject: [PATCH 2/8] =?UTF-8?q?0236=E4=BA=8C=E5=8F=89=E6=A0=91=E7=9A=84?= =?UTF-8?q?=E6=9C=80=E8=BF=91=E5=85=AC=E5=85=B1=E7=A5=96=E5=85=88JavaScrip?= =?UTF-8?q?t=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../0236.二叉树的最近公共祖先.md | 26 ++++++++++++++++++- 1 file changed, 25 insertions(+), 1 deletion(-) diff --git a/problems/0236.二叉树的最近公共祖先.md b/problems/0236.二叉树的最近公共祖先.md index 2f7aa6c3..7673a0ab 100644 --- a/problems/0236.二叉树的最近公共祖先.md +++ b/problems/0236.二叉树的最近公共祖先.md @@ -310,7 +310,31 @@ func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode { return nil } ``` - +JavaScript版本： +```javascript +var lowestCommonAncestor = function(root, p, q) { + // 使用递归的方法 + // 需要从下到上，所以使用后序遍历 + // 1. 确定递归的函数 + const travelTree = function(root,p,q) { + // 2. 确定递归终止条件 + if(root === null || root === p||root === q) { + return root; + } + // 3. 确定递归单层逻辑 + let left = travelTree(root.left,p,q); + let right = travelTree(root.right,p,q); + if(left !== null&&right !== null) { + return root; + } + if(left ===null) { + return right; + } + return left; + } + return travelTree(root,p,q); +}; +``` ----------------------- From 3dc4167c59a29b7654d0a03a76645bee1e6e57db Mon Sep 17 00:00:00 2001 From: Yan Wen Date: Sun, 27 Jun 2021 09:19:45 +0800 Subject: [PATCH 3/8] =?UTF-8?q?update=E5=B7=A6=E6=97=8B=E8=BD=AC=E5=AD=97?= =?UTF-8?q?=E7=AC=A6=E4=B8=B2=EF=BC=8C=E6=B7=BB=E5=8A=A0python=E4=BB=A3?= =?UTF-8?q?=E7=A0=81?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../剑指Offer58-II.左旋转字符串.md | 21 +++++++++++++++++++ 1 file changed, 21 insertions(+) diff --git a/problems/剑指Offer58-II.左旋转字符串.md b/problems/剑指Offer58-II.左旋转字符串.md index 39c8382c..ca903900 100644 --- a/problems/剑指Offer58-II.左旋转字符串.md +++ b/problems/剑指Offer58-II.左旋转字符串.md @@ -119,6 +119,27 @@ class Solution { ``` Python： +```python +# 方法一：可以使用切片方法 +class Solution: + def reverseLeftWords(self, s: str, n: int) -> str: + return s[n:] + s[0:n] + +# 方法二：也可以使用上文描述的方法，有些面试中不允许使用切片，那就使用上文作者提到的方法 +# class Solution: +# def reverseLeftWords(self, s: str, n: int) -> str: +# s = list(s) +# s[0:n] = list(reversed(s[0:n])) +# s[n:] = list(reversed(s[n:])) +# s.reverse() + +# return "".join(s) + + +# 时间复杂度：O(n) +# 空间复杂度：O(n)，python的string为不可变，需要开辟同样大小的list空间来修改 +``` + Go： ```go From c0dbcc8a73eec6517b8b26e376a8c7060b8b858b Mon Sep 17 00:00:00 2001 From: "xeniaxie(xl)" Date: Sun, 27 Jun 2021 10:28:58 +0800 Subject: [PATCH 4/8] =?UTF-8?q?0235=E4=BA=8C=E5=8F=89=E6=90=9C=E7=B4=A2?= =?UTF-8?q?=E6=A0=91=E7=9A=84=E6=9C=80=E8=BF=91=E5=85=AC=E5=85=B1=E7=A5=96?= =?UTF-8?q?=E5=85=88?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- ...35.二叉搜索树的最近公共祖先.md | 41 ++++++++++++++++++- 1 file changed, 40 insertions(+), 1 deletion(-) diff --git a/problems/0235.二叉搜索树的最近公共祖先.md b/problems/0235.二叉搜索树的最近公共祖先.md index d78db42a..b8b23f09 100644 --- a/problems/0235.二叉搜索树的最近公共祖先.md +++ b/problems/0235.二叉搜索树的最近公共祖先.md @@ -312,7 +312,46 @@ func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode { }else {return findLeft} } ``` - +JavaScript版本： +1. 使用递归的方法： +```javascript +var lowestCommonAncestor = function(root, p, q) { + // 使用递归的方法 + // 1. 使用给定的递归函数lowestCommonAncestor + // 2. 确定递归终止条件 + if(root === null) { + return root; + } + if(root.val>p.val&&root.val>q.val) { + // 向左子树查询 + let left = lowestCommonAncestor(root.left,p,q); + return left !== null&&left; + } + if(root.valp.val&&root.val>q.val) { + root = root.left; + }else if(root.val Date: Sun, 27 Jun 2021 10:35:03 +0800 Subject: [PATCH 5/8] =?UTF-8?q?Revert=20"0235=E4=BA=8C=E5=8F=89=E6=90=9C?= =?UTF-8?q?=E7=B4=A2=E6=A0=91=E7=9A=84=E6=9C=80=E8=BF=91=E5=85=AC=E5=85=B1?= =?UTF-8?q?=E7=A5=96=E5=85=88"?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit This reverts commit c0dbcc8a73eec6517b8b26e376a8c7060b8b858b. --- ...35.二叉搜索树的最近公共祖先.md | 41 +------------------ 1 file changed, 1 insertion(+), 40 deletions(-) diff --git a/problems/0235.二叉搜索树的最近公共祖先.md b/problems/0235.二叉搜索树的最近公共祖先.md index b8b23f09..d78db42a 100644 --- a/problems/0235.二叉搜索树的最近公共祖先.md +++ b/problems/0235.二叉搜索树的最近公共祖先.md @@ -312,46 +312,7 @@ func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode { }else {return findLeft} } ``` -JavaScript版本： -1. 使用递归的方法： -```javascript -var lowestCommonAncestor = function(root, p, q) { - // 使用递归的方法 - // 1. 使用给定的递归函数lowestCommonAncestor - // 2. 确定递归终止条件 - if(root === null) { - return root; - } - if(root.val>p.val&&root.val>q.val) { - // 向左子树查询 - let left = lowestCommonAncestor(root.left,p,q); - return left !== null&&left; - } - if(root.valp.val&&root.val>q.val) { - root = root.left; - }else if(root.val Date: Sun, 27 Jun 2021 10:45:48 +0800 Subject: [PATCH 6/8] =?UTF-8?q?0235=E4=BA=8C=E5=8F=89=E6=90=9C=E7=B4=A2?= =?UTF-8?q?=E6=A0=91=E7=9A=84=E6=9C=80=E8=BF=91=E5=85=AC=E5=85=B1=E7=A5=96?= =?UTF-8?q?=E5=85=88Javascript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- ...35.二叉搜索树的最近公共祖先.md | 42 ++++++++++++++++++- 1 file changed, 40 insertions(+), 2 deletions(-) diff --git a/problems/0235.二叉搜索树的最近公共祖先.md b/problems/0235.二叉搜索树的最近公共祖先.md index d78db42a..8ea3cdc5 100644 --- a/problems/0235.二叉搜索树的最近公共祖先.md +++ b/problems/0235.二叉搜索树的最近公共祖先.md @@ -312,8 +312,46 @@ func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode { }else {return findLeft} } ``` - - +JavaScript版本： +1. 使用递归的方法 +```javascript +var lowestCommonAncestor = function(root, p, q) { + // 使用递归的方法 + // 1. 使用给定的递归函数lowestCommonAncestor + // 2. 确定递归终止条件 + if(root === null) { + return root; + } + if(root.val>p.val&&root.val>q.val) { + // 向左子树查询 + let left = lowestCommonAncestor(root.left,p,q); + return left !== null&&left; + } + if(root.valp.val&&root.val>q.val) { + root = root.left; + }else if(root.val Date: Sun, 27 Jun 2021 13:11:31 +0800 Subject: [PATCH 7/8] =?UTF-8?q?Update=20=E8=83=8C=E5=8C=85=E7=90=86?= =?UTF-8?q?=E8=AE=BA=E5=9F=BA=E7=A1=8001=E8=83=8C=E5=8C=85-1.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/背包理论基础01背包-1.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/背包理论基础01背包-1.md b/problems/背包理论基础01背包-1.md index 1269d9c1..3cbfb347 100644 --- a/problems/背包理论基础01背包-1.md +++ b/problems/背包理论基础01背包-1.md @@ -82,7 +82,7 @@ leetcode上没有纯01背包的问题，都是01背包应用方面的题目， 那么可以有两个方向推出来dp[i][j]， -* 由dp[i - 1][j]推出，即背包容量为j，里面不放物品i的最大价值，此时dp[i][j]就是dp[i - 1][j] +* 由dp[i - 1][j]推出，即背包容量为j，里面不放物品i的最大价值，此时dp[i][j]就是dp[i - 1][j]。(其实就是当物品i的重量大于背包j的重量时，物品i无法放进背包中，所以被背包内的价值依然和前面相同。) * 由dp[i - 1][j - weight[i]]推出，dp[i - 1][j - weight[i]] 为背包容量为j - weight[i]的时候不放物品i的最大价值，那么dp[i - 1][j - weight[i]] + value[i] （物品i的价值），就是背包放物品i得到的最大价值 所以递归公式： dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - weight[i]] + value[i]); From 5e68f30728cbcdb670eda7f12d3c208796f5c731 Mon Sep 17 00:00:00 2001 From: Jijie LIU Date: Sun, 27 Jun 2021 17:40:11 +0200 Subject: [PATCH 8/8] =?UTF-8?q?update=20279.=E5=AE=8C=E5=85=A8=E5=B9=B3?= =?UTF-8?q?=E6=96=B9=E6=95=B0:=20=E6=8F=90=E4=BE=9B=E6=96=B0=E7=9A=84?= =?UTF-8?q?=E5=88=9D=E5=A7=8B=E5=8C=96=E6=96=B9=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0279.完全平方数.md | 18 ++++++++++++++++++ 1 file changed, 18 insertions(+) diff --git a/problems/0279.完全平方数.md b/problems/0279.完全平方数.md index 60d6d165..d0922de1 100644 --- a/problems/0279.完全平方数.md +++ b/problems/0279.完全平方数.md @@ -214,8 +214,26 @@ class Solution: return dp[n] ``` +Python3: +```python +class Solution: + def numSquares(self, n: int) -> int: + # 初始化 + # 组成和的完全平方数的最多个数，就是只用1构成 + # 因此，dp[i] = i + dp = [i for i in range(n + 1)] + # dp[0] = 0 无意义，只是为了方便记录特殊情况: + # n本身就是完全平方数，dp[n] = min(dp[n], dp[n - n] + 1) = 1 + for i in range(1, n): # 遍历物品 + if i * i > n: + break + num = i * i + for j in range(num, n + 1): # 遍历背包 + dp[j] = min(dp[j], dp[j - num] + 1) + return dp[n] +``` Go： ```go