0047-全排列II

给定一个可包含重复数字的序列 nums ，按任意顺序 返回所有不重复的全排列。

示例 1：

输入：nums = [1,1,2]
输出：
[[1,1,2],
[1,2,1],
[2,1,1]]
示例 2：

输入：nums = [1,2,3]
输出：[[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]

提示：

1 <= nums.length <= 8
-10 <= nums[i] <= 10

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/permutations-ii

参考：

• https://leetcode-cn.com/problems/permutations-ii/solution/dai-ma-sui-xiang-lu-dai-ni-xue-tou-hui-s-ki1h/

python

# 0047.全排列II
 
class Solution:
    def permuteUnique(self, nums: [int]) -> [[int]]:
        if not nums:
            return []
        res = []
        used = [0] * len(nums)
 
        def track(nums, used, path):
            if len(path) == len(nums):
                res.append(path.copy())
                return
            for i in range(len(nums)):
                if not used[i]:
                    if i > 0 and nums[i] == nums[i-1] and not used[i-1]:
                        continue
                    used[i] = 1
                    path.append(nums[i])
                    track(nums, used, path)
                    path.pop()
                    used[i] = 0
 
        nums = sorted(nums)
        track(nums, used, [])
        return res

golang

package backTrack
 
var res [][]int
func permuteUnique(nums []int) [][]int {
	res = [][]int{}
	backTrack(nums,len(nums),[]int{})
	return res
}
func backTrack(nums []int,numsLen int,path []int)  {
	if len(nums)==0{
		p:=make([]int,len(path))
		copy(p,path)
		res = append(res,p)
	}
	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
	for i:=0;i<numsLen;i++{
		if used[nums[i]+10]==1{
			continue
		}
		cur:=nums[i]
		path = append(path,cur)
		used[nums[i]+10]=1
		nums = append(nums[:i],nums[i+1:]...)
		backTrack(nums,len(nums),path)
		nums = append(nums[:i],append([]int{cur},nums[i:]...)...)
		path = path[:len(path)-1]
	}
}