From 3aa54bfb8a8604a8cfb05a4fa8c606ebe6b09db3 Mon Sep 17 00:00:00 2001 From: Wayne <3522373084@qq.com> Date: Mon, 7 Feb 2022 21:39:51 +0800 Subject: [PATCH 1/5] =?UTF-8?q?343=E6=95=B4=E6=95=B0=E6=8B=86=E5=88=86,?= =?UTF-8?q?=E6=9B=B4=E6=96=B0=E7=AC=AC=E4=BA=8C=E5=B1=82=E5=BE=AA=E7=8E=AF?= =?UTF-8?q?=20j=20=E7=9A=84=E8=8C=83=E5=9B=B4?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0343.整数拆分.md | 19 +++++++++++-------- 1 file changed, 11 insertions(+), 8 deletions(-) diff --git a/problems/0343.整数拆分.md b/problems/0343.整数拆分.md index 5d11f670..f616a606 100644 --- a/problems/0343.整数拆分.md +++ b/problems/0343.整数拆分.md @@ -197,14 +197,17 @@ Java： ```Java class Solution { public int integerBreak(int n) { - //dp[i]为正整数i拆分结果的最大乘积 - int[] dp = new int[n+1]; - dp[2] = 1; - for (int i = 3; i <= n; ++i) { - for (int j = 1; j < i - 1; ++j) { - //j*(i-j)代表把i拆分为j和i-j两个数相乘 - //j*dp[i-j]代表把i拆分成j和继续把(i-j)这个数拆分，取(i-j)拆分结果中的最大乘积与j相乘 - dp[i] = Math.max(dp[i], Math.max(j * (i - j), j * dp[i - j])); + //dp[i] 为正整数 i 拆分后的结果的最大乘积 + int[]dp=new int[n+1]; + dp[2]=1; + for(int i=3;i<=n;i++){ + for(int j=1;j<=i-j;j++){ + // 这里的 j 其实最大值为 i-j,再大只不过是重复而已， + //并且，在本题中，我们分析 dp[0], dp[1]都是无意义的， + //j 最大到 i-j,就不会用到 dp[0]与dp[1] + dp[i]=Math.max(dp[i],Math.max(j*(i-j),j*dp[i-j])); + // j * (i - j) 是单纯的把整数 i 拆分为两个数 也就是 i,i-j ，再相乘 + //而j * dp[i - j]是将 i 拆分成两个以及两个以上的个数,再相乘。 } } return dp[n]; From 1b7c86e20c2b36ed0732deb66a0c124a8e8fcdff Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Mon, 7 Feb 2022 23:01:47 +0800 Subject: [PATCH 2/5] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880257.=E4=BA=8C?= =?UTF-8?q?=E5=8F=89=E6=A0=91=E7=9A=84=E6=89=80=E6=9C=89=E8=B7=AF=E5=BE=84?= =?UTF-8?q?.md=EF=BC=89=EF=BC=9A=E5=A2=9E=E5=8A=A0typescript=E7=89=88?= =?UTF-8?q?=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0257.二叉树的所有路径.md | 67 +++++++++++++++++++++-- 1 file changed, 61 insertions(+), 6 deletions(-) diff --git a/problems/0257.二叉树的所有路径.md b/problems/0257.二叉树的所有路径.md index 4078320f..1362897c 100644 --- a/problems/0257.二叉树的所有路径.md +++ b/problems/0257.二叉树的所有路径.md @@ -433,9 +433,9 @@ class Solution: if cur.right: self.traversal(cur.right, path + '->', result) ``` - + 迭代法： - + ```python3 from collections import deque @@ -463,13 +463,13 @@ class Solution: return result ``` - + --- Go： - + 递归法： - + ```go func binaryTreePaths(root *TreeNode) []string { res := make([]string, 0) @@ -492,7 +492,7 @@ func binaryTreePaths(root *TreeNode) []string { return res } ``` - + 迭代法： ```go @@ -581,7 +581,62 @@ var binaryTreePaths = function(root) { }; ``` +TypeScript： + +> 递归法 + +```typescript +function binaryTreePaths(root: TreeNode | null): string[] { + function recur(node: TreeNode, route: string, resArr: string[]): void { + route += String(node.val); + if (node.left === null && node.right === null) { + resArr.push(route); + return; + } + if (node.left !== null) recur(node.left, route + '->', resArr); + if (node.right !== null) recur(node.right, route + '->', resArr); + } + const resArr: string[] = []; + if (root === null) return resArr; + recur(root, '', resArr); + return resArr; +}; +``` + +> 迭代法 + +```typescript +// 迭代法2 +function binaryTreePaths(root: TreeNode | null): string[] { + let helperStack: TreeNode[] = []; + let tempNode: TreeNode; + let routeArr: string[] = []; + let resArr: string[] = []; + if (root !== null) { + helperStack.push(root); + routeArr.push(String(root.val)); + }; + while (helperStack.length > 0) { + tempNode = helperStack.pop()!; + let route: string = routeArr.pop()!; // tempNode 对应的路径 + if (tempNode.left === null && tempNode.right === null) { + resArr.push(route); + } + if (tempNode.right !== null) { + helperStack.push(tempNode.right); + routeArr.push(route + '->' + tempNode.right.val); // tempNode.right 对应的路径 + } + if (tempNode.left !== null) { + helperStack.push(tempNode.left); + routeArr.push(route + '->' + tempNode.left.val); // tempNode.left 对应的路径 + } + } + return resArr; +}; +``` + Swift: + > 递归/回溯 ```swift func binaryTreePaths(_ root: TreeNode?) -> [String] { From 47818c947888ecff3dae6f36daded1cfaf92db3b Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Tue, 8 Feb 2022 10:23:59 +0800 Subject: [PATCH 3/5] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=88=E4=BA=8C?= =?UTF-8?q?=E5=8F=89=E6=A0=91=E4=B8=AD=E9=80=92=E5=BD=92=E5=B8=A6=E7=9D=80?= =?UTF-8?q?=E5=9B=9E=E6=BA=AF.md=EF=BC=89=EF=BC=9A=E5=A2=9E=E5=8A=A0?= =?UTF-8?q?=E7=9B=B8=E5=90=8C=E7=9A=84=E6=A0=91typescript=E7=89=88?= =?UTF-8?q?=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/二叉树中递归带着回溯.md | 23 ++++++++++++++++++++++ 1 file changed, 23 insertions(+) diff --git a/problems/二叉树中递归带着回溯.md b/problems/二叉树中递归带着回溯.md index 20b87f87..03815ed3 100644 --- a/problems/二叉树中递归带着回溯.md +++ b/problems/二叉树中递归带着回溯.md @@ -515,6 +515,29 @@ var binaryTreePaths = function(root) { }; ``` +TypeScript： + +> 相同的树 + +```typescript +function isSameTree(p: TreeNode | null, q: TreeNode | null): boolean { + if (p === null && q === null) return true; + if (p === null || q === null) return false; + if (p.val !== q.val) return false; + let bool1: boolean, bool2: boolean; + bool1 = isSameTree(p.left, q.left); + bool2 = isSameTree(p.right, q.right); + return bool1 && bool2; +}; +``` + +> 二叉树的不同路径 + +```typescript +``` + + + -----------------------
From 49a1a42bbed494b10b4d6659ba787830f62bd842 Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Tue, 8 Feb 2022 10:45:15 +0800 Subject: [PATCH 4/5] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=88=E4=BA=8C?= =?UTF-8?q?=E5=8F=89=E6=A0=91=E4=B8=AD=E9=80=92=E5=BD=92=E5=B8=A6=E7=9D=80?= =?UTF-8?q?=E5=9B=9E=E6=BA=AF.md=EF=BC=89=EF=BC=9A=E5=A2=9E=E5=8A=A0?= =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=A0=91=E7=9A=84=E4=B8=8D=E5=90=8C=E8=B7=AF?= =?UTF-8?q?=E5=BE=84=E7=9A=84typescript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/二叉树中递归带着回溯.md | 21 +++++++++++++++++++++ 1 file changed, 21 insertions(+) diff --git a/problems/二叉树中递归带着回溯.md b/problems/二叉树中递归带着回溯.md index 03815ed3..41d5663e 100644 --- a/problems/二叉树中递归带着回溯.md +++ b/problems/二叉树中递归带着回溯.md @@ -534,6 +534,27 @@ function isSameTree(p: TreeNode | null, q: TreeNode | null): boolean { > 二叉树的不同路径 ```typescript +function binaryTreePaths(root: TreeNode | null): string[] { + function recur(node: TreeNode, nodeSeqArr: number[], resArr: string[]): void { + nodeSeqArr.push(node.val); + if (node.left === null && node.right === null) { + resArr.push(nodeSeqArr.join('->')); + } + if (node.left !== null) { + recur(node.left, nodeSeqArr, resArr); + nodeSeqArr.pop(); + } + if (node.right !== null) { + recur(node.right, nodeSeqArr, resArr); + nodeSeqArr.pop(); + } + } + let nodeSeqArr: number[] = []; + let resArr: string[] = []; + if (root === null) return resArr; + recur(root, nodeSeqArr, resArr); + return resArr; +}; ``` From f604aea34c6632e20ad4ced37cb24c005420e574 Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Tue, 8 Feb 2022 23:00:31 +0800 Subject: [PATCH 5/5] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880404.=E5=B7=A6?= =?UTF-8?q?=E5=8F=B6=E5=AD=90=E4=B9=8B=E5=92=8C.md=EF=BC=89=EF=BC=9A?= =?UTF-8?q?=E5=A2=9E=E5=8A=A0typescript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0404.左叶子之和.md | 44 ++++++++++++++++++++++++++++++++ 1 file changed, 44 insertions(+) diff --git a/problems/0404.左叶子之和.md b/problems/0404.左叶子之和.md index 09272052..2b4df151 100644 --- a/problems/0404.左叶子之和.md +++ b/problems/0404.左叶子之和.md @@ -372,6 +372,50 @@ var sumOfLeftLeaves = function(root) { }; ``` +## TypeScript + +> 递归法 + +```typescript +function sumOfLeftLeaves(root: TreeNode | null): number { + if (root === null) return 0; + let midVal: number = 0; + if ( + root.left !== null && + root.left.left === null && + root.left.right === null + ) { + midVal = root.left.val; + } + let leftVal: number = sumOfLeftLeaves(root.left); + let rightVal: number = sumOfLeftLeaves(root.right); + return midVal + leftVal + rightVal; +}; +``` + +> 迭代法 + +```typescript +function sumOfLeftLeaves(root: TreeNode | null): number { + let helperStack: TreeNode[] = []; + let tempNode: TreeNode; + let sum: number = 0; + if (root !== null) helperStack.push(root); + while (helperStack.length > 0) { + tempNode = helperStack.pop()!; + if ( + tempNode.left !== null && + tempNode.left.left === null && + tempNode.left.right === null + ) { + sum += tempNode.left.val; + } + if (tempNode.right !== null) helperStack.push(tempNode.right); + if (tempNode.left !== null) helperStack.push(tempNode.left); + } + return sum; +}; +``` ## Swift