47. 全排列 II @Python @Go

原题

https://leetcode.cn/problems/permutations-ii/description/

思路

回溯法 + 去重

复杂度

时间：O(C(n, n))
空间：O(C(n, n))

Python代码

class Solution:
    def permuteUnique(self, nums: List[int]) -> List[List[int]]:
        def backtrack(path):
            if len(path) == len(nums):
                ans.append(path.copy())
                return
            for i in range(len(nums)):
                # 去重
                if visited[i] or (
                    i > 0 and nums[i] == nums[i - 1] and not visited[i - 1]
                ):
                    continue
                visited[i] = True
                path.append(nums[i])
                # 回溯
                backtrack(path)
                # 还原
                path.pop()
                visited[i] = False

        ans = []
        # 排序
        nums.sort()
        visited = [False] * len(nums)
        backtrack([])
        return ans

Go代码

func permuteUnique(nums []int) [][]int {
	var ans [][]int
	// 排序
	sort.Ints(nums)
	visited := make([]bool, len(nums))
	// 匿名函数
	var backtrack func([]int)
	backtrack = func(path []int) {
		if len(path) == len(nums) {
			// 复制路径
			com := make([]int, len(path))
			copy(com, path)
			ans = append(ans, com)
			return
		}
		for i := 0; i < len(nums); i++ {
			// 去重
			if visited[i] || (i > 0 && nums[i] == nums[i-1] && !visited[i-1]) {
				continue
			}
			visited[i] = true
			path = append(path, nums[i])
			// 回溯
			backtrack(path)
			// 还原
			path = path[:len(path)-1]
			visited[i] = false
		}
	}

	backtrack([]int{})
	return ans
}