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update 0047.全排列II:替换 go 代码,改错字
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@ -49,7 +49,7 @@
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**一般来说:组合问题和排列问题是在树形结构的叶子节点上收集结果,而子集问题就是取树上所有节点的结果**。
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在[46.全排列](https://programmercarl.com/0046.全排列.html)中已经详解讲解了排列问题的写法,在[40.组合总和II](https://programmercarl.com/0040.组合总和II.html) 、[90.子集II](https://programmercarl.com/0090.子集II.html)中详细讲解的去重的写法,所以这次我就不用回溯三部曲分析了,直接给出代码,如下:
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在[46.全排列](https://programmercarl.com/0046.全排列.html)中已经详细讲解了排列问题的写法,在[40.组合总和II](https://programmercarl.com/0040.组合总和II.html) 、[90.子集II](https://programmercarl.com/0090.子集II.html)中详细讲解了去重的写法,所以这次我就不用回溯三部曲分析了,直接给出代码,如下:
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## C++代码
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@ -225,33 +225,37 @@ class Solution:
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### Go
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```go
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var res [][]int
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func permute(nums []int) [][]int {
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res = [][]int{}
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backTrack(nums,len(nums),[]int{})
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return res
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var (
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res [][]int
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path []int
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st []bool // state的缩写
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)
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func permuteUnique(nums []int) [][]int {
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res, path = make([][]int, 0), make([]int, 0, len(nums))
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st = make([]bool, len(nums))
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sort.Ints(nums)
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dfs(nums, 0)
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return res
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}
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func backTrack(nums []int,numsLen int,path []int) {
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if len(nums)==0{
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p:=make([]int,len(path))
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copy(p,path)
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res = append(res,p)
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}
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used := [21]int{}//跟前一题唯一的区别,同一层不使用重复的数。关于used的思想carl在递增子序列那一题中提到过
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for i:=0;i<numsLen;i++{
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if used[nums[i]+10]==1{
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continue
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}
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cur:=nums[i]
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path = append(path,cur)
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used[nums[i]+10]=1
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nums = append(nums[:i],nums[i+1:]...)
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backTrack(nums,len(nums),path)
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nums = append(nums[:i],append([]int{cur},nums[i:]...)...)
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path = path[:len(path)-1]
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}
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func dfs(nums []int, cur int) {
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if cur == len(nums) {
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tmp := make([]int, len(path))
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copy(tmp, path)
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res = append(res, tmp)
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}
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for i := 0; i < len(nums); i++ {
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if i != 0 && nums[i] == nums[i-1] && !st[i-1] { // 去重,用st来判别是深度还是广度
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continue
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}
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if !st[i] {
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path = append(path, nums[i])
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st[i] = true
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dfs(nums, cur + 1)
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st[i] = false
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path = path[:len(path)-1]
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}
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}
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}
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```
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