From fb67fb4320c3ee7c0ee17dd695a5b7a7ea2a4d34 Mon Sep 17 00:00:00 2001 From: "darion.yaphet" Date: Sat, 12 Feb 2022 17:25:57 +0800 Subject: [PATCH 1/7] fix java code --- problems/背包理论基础01背包-1.md | 8 ++++---- 1 file changed, 4 insertions(+), 4 deletions(-) diff --git a/problems/背包理论基础01背包-1.md b/problems/背包理论基础01背包-1.md index 6ff32017..fe940b4c 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 @@ -271,7 +271,7 @@ int main() { ### java ```java - public static void main(string[] args) { + public static void main(String[] args) { int[] weight = {1, 3, 4}; int[] value = {15, 20, 30}; int bagsize = 4; @@ -292,16 +292,16 @@ int main() { if (j < weight[i - 1]){ dp[i][j] = dp[i - 1][j]; }else{ - dp[i][j] = math.max(dp[i - 1][j], dp[i - 1][j - weight[i - 1]] + value[i - 1]); + dp[i][j] = Math.max(dp[i - 1][j], dp[i - 1][j - weight[i - 1]] + value[i - 1]); } } } //打印dp数组 for (int i = 0; i <= wlen; i++){ for (int j = 0; j <= bagsize; j++){ - system.out.print(dp[i][j] + " "); + System.out.print(dp[i][j] + " "); } - system.out.print("\n"); + System.out.print("\n"); } } ``` From bac5f8d70cef72f5457ac80fd487d86618d27831 Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Sun, 13 Feb 2022 16:37:32 +0800 Subject: [PATCH 2/7] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880106.=E4=BB=8E?= =?UTF-8?q?=E4=B8=AD=E5=BA=8F=E4=B8=8E=E5=90=8E=E5=BA=8F=E9=81=8D=E5=8E=86?= =?UTF-8?q?=E5=BA=8F=E5=88=97=E6=9E=84=E9=80=A0=E4=BA=8C=E5=8F=89=E6=A0=91?= =?UTF-8?q?.md=EF=BC=89=EF=BC=9A=E5=A2=9E=E5=8A=A0typescript=E7=89=88?= =?UTF-8?q?=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- ...序与后序遍历序列构造二叉树.md | 63 +++++++++++++++++++ 1 file changed, 63 insertions(+) diff --git a/problems/0106.从中序与后序遍历序列构造二叉树.md b/problems/0106.从中序与后序遍历序列构造二叉树.md index 40d75983..9b18e729 100644 --- a/problems/0106.从中序与后序遍历序列构造二叉树.md +++ b/problems/0106.从中序与后序遍历序列构造二叉树.md @@ -814,7 +814,70 @@ var buildTree = function(preorder, inorder) { }; ``` +## TypeScript + +> 106.从中序与后序遍历序列构造二叉树 + +**创建新数组:** + +```typescript +function buildTree(inorder: number[], postorder: number[]): TreeNode | null { + if (postorder.length === 0) return null; + const rootVal: number = postorder.pop()!; + const rootValIndex: number = inorder.indexOf(rootVal); + const rootNode: TreeNode = new TreeNode(rootVal); + rootNode.left = buildTree(inorder.slice(0, rootValIndex), postorder.slice(0, rootValIndex)); + rootNode.right = buildTree(inorder.slice(rootValIndex + 1), postorder.slice(rootValIndex)); + return rootNode; +}; +``` + +**使用数组索引:** + +```typescript +function buildTree(inorder: number[], postorder: number[]): TreeNode | null { + function recur( + inorder: number[], postorder: number[], + inBegin: number, inEnd: number, + postBegin: number, postEnd: number + ): TreeNode | null { + if (postBegin === postEnd) return null; + const rootVal: number = postorder[postEnd - 1]!; + const rootValIndex: number = inorder.indexOf(rootVal, inBegin); + const rootNode: TreeNode = new TreeNode(rootVal); + + const leftInorderBegin: number = inBegin; + const leftInorderEnd: number = rootValIndex; + const rightInorderBegin: number = rootValIndex + 1; + const rightInorderEnd: number = inEnd; + + const leftPostorderBegin: number = postBegin; + const leftPostorderEnd: number = postBegin + rootValIndex - inBegin; + const rightPostorderBegin: number = leftPostorderEnd; + const rightPostorderEnd: number = postEnd - 1; + + rootNode.left = recur( + inorder, postorder, + leftInorderBegin, leftInorderEnd, + leftPostorderBegin, leftPostorderEnd + ); + rootNode.right = recur( + inorder, postorder, + rightInorderBegin, rightInorderEnd, + rightPostorderBegin, rightPostorderEnd + ); + return rootNode; + } + return recur(inorder, postorder, 0, inorder.length, 0, inorder.length); +}; +``` + +> 105.从前序与中序遍历序列构造二叉树 + + + ## C + 106 从中序与后序遍历序列构造二叉树 ```c From ddf79ac04322e27c50f50be81a46407d7b054950 Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Sun, 13 Feb 2022 17:09:38 +0800 Subject: [PATCH 3/7] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=88105.=E4=BB=8E?= =?UTF-8?q?=E5=89=8D=E5=BA=8F=E4=B8=8E=E4=B8=AD=E5=BA=8F=E9=81=8D=E5=8E=86?= =?UTF-8?q?=E5=BA=8F=E5=88=97=E6=9E=84=E9=80=A0=E4=BA=8C=E5=8F=89=E6=A0=91?= =?UTF-8?q?=EF=BC=89=EF=BC=9A=E5=A2=9E=E5=8A=A0typescript=E7=89=88?= =?UTF-8?q?=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- ...序与后序遍历序列构造二叉树.md | 44 +++++++++++++++++++ 1 file changed, 44 insertions(+) diff --git a/problems/0106.从中序与后序遍历序列构造二叉树.md b/problems/0106.从中序与后序遍历序列构造二叉树.md index 9b18e729..3e4211fb 100644 --- a/problems/0106.从中序与后序遍历序列构造二叉树.md +++ b/problems/0106.从中序与后序遍历序列构造二叉树.md @@ -874,7 +874,51 @@ function buildTree(inorder: number[], postorder: number[]): TreeNode | null { > 105.从前序与中序遍历序列构造二叉树 +**新建数组：** +```typescript +function buildTree(preorder: number[], inorder: number[]): TreeNode | null { + if (preorder.length === 0) return null; + const rootVal: number = preorder[0]; + const rootNode: TreeNode = new TreeNode(rootVal); + const rootValIndex: number = inorder.indexOf(rootVal); + rootNode.left = buildTree(preorder.slice(1, rootValIndex + 1), inorder.slice(0, rootValIndex)); + rootNode.right = buildTree(preorder.slice(rootValIndex + 1), inorder.slice(rootValIndex + 1)); + return rootNode; +}; +``` + +**使用数组索引:** + +```typescript +function buildTree(preorder: number[], inorder: number[]): TreeNode | null { + function recur( + preorder: number[], inorder: number[], + preBegin: number, preEnd: number, + inBegin: number, inEnd: number + ): TreeNode | null { + if (preBegin === preEnd) return null; + const rootVal: number = preorder[preBegin]; + const rootNode: TreeNode = new TreeNode(rootVal); + const rootValIndex: number = inorder.indexOf(rootVal, inBegin); + + const leftPreBegin: number = preBegin + 1; + const leftPreEnd: number = preBegin + rootValIndex - inBegin + 1; + const leftInBegin: number = inBegin; + const leftInEnd: number = rootValIndex; + + const rightPreBegin: number = leftPreEnd; + const rightPreEnd: number = preEnd; + const rightInBegin: number = rootValIndex + 1; + const rightInEnd: number = inEnd; + + rootNode.left = recur(preorder, inorder, leftPreBegin, leftPreEnd, leftInBegin, leftInEnd); + rootNode.right = recur(preorder, inorder, rightPreBegin, rightPreEnd, rightInBegin, rightInEnd); + return rootNode; + }; + return recur(preorder, inorder, 0, preorder.length, 0, inorder.length); +}; +``` ## C From 92322870fe4664dffa3dcb8e6020626376417bb3 Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Sun, 13 Feb 2022 21:20:24 +0800 Subject: [PATCH 4/7] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880654.=E6=9C=80?= =?UTF-8?q?=E5=A4=A7=E4=BA=8C=E5=8F=89=E6=A0=91.md=EF=BC=89=EF=BC=9A?= =?UTF-8?q?=E5=A2=9E=E5=8A=A0typescript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0654.最大二叉树.md | 49 ++++++++++++++++++++++++++++++++ 1 file changed, 49 insertions(+) diff --git a/problems/0654.最大二叉树.md b/problems/0654.最大二叉树.md index 44c74f89..1c73354b 100644 --- a/problems/0654.最大二叉树.md +++ b/problems/0654.最大二叉树.md @@ -371,7 +371,56 @@ var constructMaximumBinaryTree = function (nums) { }; ``` +## TypeScript + +> 新建数组法： + +```typescript +function constructMaximumBinaryTree(nums: number[]): TreeNode | null { + if (nums.length === 0) return null; + let maxIndex: number = 0; + let maxVal: number = nums[0]; + for (let i = 1, length = nums.length; i < length; i++) { + if (nums[i] > maxVal) { + maxIndex = i; + maxVal = nums[i]; + } + } + const rootNode: TreeNode = new TreeNode(maxVal); + rootNode.left = constructMaximumBinaryTree(nums.slice(0, maxIndex)); + rootNode.right = constructMaximumBinaryTree(nums.slice(maxIndex + 1)); + return rootNode; +}; +``` + +> 使用数组索引法： + +```typescript +function constructMaximumBinaryTree(nums: number[]): TreeNode | null { + // 左闭右开区间[begin, end) + function recur(nums: number[], begin: number, end: number): TreeNode | null { + if (begin === end) return null; + let maxIndex: number = begin; + let maxVal: number = nums[begin]; + for (let i = begin + 1; i < end; i++) { + if (nums[i] > maxVal) { + maxIndex = i; + maxVal = nums[i]; + } + } + const rootNode: TreeNode = new TreeNode(maxVal); + rootNode.left = recur(nums, begin, maxIndex); + rootNode.right = recur(nums, maxIndex + 1, end); + return rootNode; + } + return recur(nums, 0, nums.length); +}; +``` + + + ## C + ```c struct TreeNode* traversal(int* nums, int left, int right) { //若左边界大于右边界，返回NULL From 2440d6a365def0caf92f041886545117ec0c1e9f Mon Sep 17 00:00:00 2001 From: Dewittt <43514251+Dewittt@users.noreply.github.com> Date: Mon, 14 Feb 2022 14:04:34 +0800 Subject: [PATCH 5/7] =?UTF-8?q?Update=20=E8=83=8C=E5=8C=85=E9=97=AE?= =?UTF-8?q?=E9=A2=98=E7=90=86=E8=AE=BA=E5=9F=BA=E7=A1=80=E5=AE=8C=E5=85=A8?= =?UTF-8?q?=E8=83=8C=E5=8C=85.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 删除多余字符 --- problems/背包问题理论基础完全背包.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/背包问题理论基础完全背包.md b/problems/背包问题理论基础完全背包.md index cea69c72..2cdfeffa 100644 --- a/problems/背包问题理论基础完全背包.md +++ b/problems/背包问题理论基础完全背包.md @@ -94,7 +94,7 @@ dp状态图如下： 看了这两个图，大家就会理解，完全背包中，两个for循环的先后循序，都不影响计算dp[j]所需要的值（这个值就是下标j之前所对应的dp[j]）。 -先遍历被背包在遍历物品，代码如下： +先遍历背包在遍历物品，代码如下： ```CPP // 先遍历背包，再遍历物品 From 30f2c6bc1763bb3cfe607707f96b1d9047a3d2b3 Mon Sep 17 00:00:00 2001 From: Verolilo <57696907+Verolilo@users.noreply.github.com> Date: Mon, 14 Feb 2022 14:15:33 +0800 Subject: [PATCH 6/7] =?UTF-8?q?=E6=9B=B4=E6=96=B00070.=E7=88=AC=E6=A5=BC?= =?UTF-8?q?=E6=A2=AF.md=20python=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 更新0070.爬楼梯.md python版本，添加了O(1)复杂度的python代码 --- problems/0070.爬楼梯.md | 28 ++++++++++++++++++++-------- 1 file changed, 20 insertions(+), 8 deletions(-) diff --git a/problems/0070.爬楼梯.md b/problems/0070.爬楼梯.md index 7f47a991..da19ea0e 100644 --- a/problems/0070.爬楼梯.md +++ b/problems/0070.爬楼梯.md @@ -256,16 +256,28 @@ public int climbStairs(int n) { ### Python ```python +# 空间复杂度为O(n)版本 class Solution: def climbStairs(self, n: int) -> int: - # dp[i]表示爬到第i级楼梯的种数， (1, 2) (2, 1)是两种不同的类型 - dp = [0] * (n + 1) - dp[0] = 1 - for i in range(n+1): - for j in range(1, 3): - if i>=j: - dp[i] += dp[i-j] - return dp[-1] + # dp[i] 为第 i 阶楼梯有多少种方法爬到楼顶 + dp=[0]*(n+1) + dp[0]=1 + dp[1]=1 + for i in range(2,n+1): + dp[i]=dp[i-1]+dp[i-2] + return dp[n] + +# 空间复杂度为O(1)版本 +class Solution: + def climbStairs(self, n: int) -> int: + dp=[0]*(n+1) + dp[0]=1 + dp[1]=1 + for i in range(2,n+1): + tmp=dp[0]+dp[1] + dp[0]=dp[1] + dp[1]=tmp + return dp[1] ``` ### Go From 6d5a0d0340798f773104f418d3db71797c5db717 Mon Sep 17 00:00:00 2001 From: Verolilo <57696907+Verolilo@users.noreply.github.com> Date: Wed, 16 Feb 2022 10:22:19 +0800 Subject: [PATCH 7/7] =?UTF-8?q?=E6=9B=B4=E6=96=B00213.=E6=89=93=E5=AE=B6?= =?UTF-8?q?=E5=8A=AB=E8=88=8DII.md=20=E6=9B=B4=E7=AE=80=E6=B4=81=E7=9A=84p?= =?UTF-8?q?ython=20=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0213.打家劫舍II.md | 32 +++++++++++++++++--------------- 1 file changed, 17 insertions(+), 15 deletions(-) diff --git a/problems/0213.打家劫舍II.md b/problems/0213.打家劫舍II.md index b6c2d080..8e569e46 100644 --- a/problems/0213.打家劫舍II.md +++ b/problems/0213.打家劫舍II.md @@ -123,22 +123,24 @@ Python： ```Python class Solution: def rob(self, nums: List[int]) -> int: - if (n := len(nums)) == 0: - return 0 - if n == 1: - return nums[0] - result1 = self.robRange(nums, 0, n - 2) - result2 = self.robRange(nums, 1, n - 1) - return max(result1 , result2) + #在198入门级的打家劫舍问题上分两种情况考虑 + #一是不偷第一间房，二是不偷最后一间房 + if len(nums)==1:#题目中提示nums.length>=1,所以不需要考虑len(nums)==0的情况 + return nums[0] + val1=self.roblist(nums[1:])#不偷第一间房 + val2=self.roblist(nums[:-1])#不偷最后一间房 + return max(val1,val2) - def robRange(self, nums: List[int], start: int, end: int) -> int: - if end == start: return nums[start] - dp = [0] * len(nums) - dp[start] = nums[start] - dp[start + 1] = max(nums[start], nums[start + 1]) - for i in range(start + 2, end + 1): - dp[i] = max(dp[i -2] + nums[i], dp[i - 1]) - return dp[end] + def robRange(self,nums): + l=len(nums) + dp=[0]*l + dp[0]=nums[0] + for i in range(1,l): + if i==1: + dp[i]=max(dp[i-1],nums[i]) + else: + dp[i]=max(dp[i-1],dp[i-2]+nums[i]) + return dp[-1] ``` javascipt: