From 89c9044bf984ad638d07932f3b9ebe91e2e7589d Mon Sep 17 00:00:00 2001 From: xiaojun <13589818805@163.com> Date: Thu, 23 Jun 2022 15:36:55 +0800 Subject: [PATCH 1/2] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=881382.=20?= =?UTF-8?q?=E5=B0=86=E4=BA=8C=E5=8F=89=E6=90=9C=E7=B4=A2=E6=A0=91=E5=8F=98?= =?UTF-8?q?=E5=B9=B3=E8=A1=A1=EF=BC=89=E7=9A=84go=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/1382.将二叉搜索树变平衡.md | 40 ++++++++++++++++++++ 1 file changed, 40 insertions(+) diff --git a/problems/1382.将二叉搜索树变平衡.md b/problems/1382.将二叉搜索树变平衡.md index 57231ec4..7c7f8484 100644 --- a/problems/1382.将二叉搜索树变平衡.md +++ b/problems/1382.将二叉搜索树变平衡.md @@ -123,6 +123,46 @@ class Solution: ``` Go： +```go +/** + * Definition for a binary tree node. + * type TreeNode struct { + * Val int + * Left *TreeNode + * Right *TreeNode + * } + */ +func balanceBST(root *TreeNode) *TreeNode { + // 二叉搜索树中序遍历得到有序数组 + nums := []int{} + // 中序递归遍历二叉树 + var travel func(node *TreeNode) + travel = func(node *TreeNode) { + if node == nil { + return + } + travel(node.Left) + nums = append(nums, node.Val) + travel(node.Right) + } + // 二分法保证左右子树高度差不超过一（题目要求返回的仍是二叉搜索树） + var buildTree func(nums []int, left, right int) *TreeNode + buildTree = func(nums []int, left, right int) *TreeNode { + if left > right { + return nil + } + mid := left + (right-left) >> 1 + root := &TreeNode{Val: nums[mid]} + root.Left = buildTree(nums, left, mid-1) + root.Right = buildTree(nums, mid+1, right) + return root + } + travel(root) + return buildTree(nums, 0, len(nums)-1) +} + +``` + JavaScript： ```javascript var balanceBST = function(root) { From ca00ec680557689d6b302fe27b3e66e479a0d421 Mon Sep 17 00:00:00 2001 From: xiaojun <13589818805@163.com> Date: Mon, 4 Jul 2022 15:43:35 +0800 Subject: [PATCH 2/2] =?UTF-8?q?=E6=B7=BB=E5=8A=A0(0031.=E4=B8=8B=E4=B8=80?= =?UTF-8?q?=E4=B8=AA=E6=8E=92=E5=88=97.md)go=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0031.下一个排列.md | 24 ++++++++++++++++++++++-- 1 file changed, 22 insertions(+), 2 deletions(-) diff --git a/problems/0031.下一个排列.md b/problems/0031.下一个排列.md index bce8adef..1a3641b0 100644 --- a/problems/0031.下一个排列.md +++ b/problems/0031.下一个排列.md @@ -171,13 +171,13 @@ class Solution(object): i = n-2 while i >= 0 and nums[i] >= nums[i+1]: i -= 1 - + if i > -1: // i==-1,不存在下一个更大的排列 j = n-1 while j >= 0 and nums[j] <= nums[i]: j -= 1 nums[i], nums[j] = nums[j], nums[i] - + start, end = i+1, n-1 while start < end: nums[start], nums[end] = nums[end], nums[start] @@ -190,6 +190,26 @@ class Solution(object): ## Go ```go +//卡尔的解法 +func nextPermutation(nums []int) { + for i:=len(nums)-1;i>=0;i--{ + for j:=len(nums)-1;j>i;j--{ + if nums[j]>nums[i]{ + //交换 + nums[j],nums[i]=nums[i],nums[j] + reverse(nums,0+i+1,len(nums)-1) + return + } + } + } + reverse(nums,0,len(nums)-1) +} +//对目标切片指定区间的反转方法 +func reverse(a []int,begin,end int){ + for i,j:=begin,end;i