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https://github.com/youngyangyang04/leetcode-master.git
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Merge pull request #1491 from jujunwang/master
添加(1382. 将二叉搜索树变平衡 0031.下一个排列)的go版本
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@ -171,13 +171,13 @@ class Solution(object):
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i = n-2
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while i >= 0 and nums[i] >= nums[i+1]:
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i -= 1
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if i > -1: // i==-1,不存在下一个更大的排列
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j = n-1
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while j >= 0 and nums[j] <= nums[i]:
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j -= 1
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nums[i], nums[j] = nums[j], nums[i]
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start, end = i+1, n-1
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while start < end:
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nums[start], nums[end] = nums[end], nums[start]
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@ -190,6 +190,26 @@ class Solution(object):
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## Go
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```go
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//卡尔的解法
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func nextPermutation(nums []int) {
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for i:=len(nums)-1;i>=0;i--{
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for j:=len(nums)-1;j>i;j--{
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if nums[j]>nums[i]{
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//交换
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nums[j],nums[i]=nums[i],nums[j]
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reverse(nums,0+i+1,len(nums)-1)
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return
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}
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}
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}
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reverse(nums,0,len(nums)-1)
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}
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//对目标切片指定区间的反转方法
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func reverse(a []int,begin,end int){
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for i,j:=begin,end;i<j;i,j=i+1,j-1{
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a[i],a[j]=a[j],a[i]
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}
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}
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```
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## JavaScript
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@ -123,6 +123,46 @@ class Solution:
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```
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Go:
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```go
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/**
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* Definition for a binary tree node.
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* type TreeNode struct {
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* Val int
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* Left *TreeNode
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* Right *TreeNode
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* }
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*/
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func balanceBST(root *TreeNode) *TreeNode {
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// 二叉搜索树中序遍历得到有序数组
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nums := []int{}
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// 中序递归遍历二叉树
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var travel func(node *TreeNode)
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travel = func(node *TreeNode) {
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if node == nil {
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return
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}
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travel(node.Left)
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nums = append(nums, node.Val)
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travel(node.Right)
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}
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// 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)
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var buildTree func(nums []int, left, right int) *TreeNode
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buildTree = func(nums []int, left, right int) *TreeNode {
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if left > right {
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return nil
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}
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mid := left + (right-left) >> 1
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root := &TreeNode{Val: nums[mid]}
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root.Left = buildTree(nums, left, mid-1)
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root.Right = buildTree(nums, mid+1, right)
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return root
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}
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travel(root)
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return buildTree(nums, 0, len(nums)-1)
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}
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```
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JavaScript:
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```javascript
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var balanceBST = function(root) {
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