0046全排列 Marathon

给定一个不含重复数字的数组 nums ，返回其 所有可能的全排列 。你可以 按任意顺序 返回答案。

示例 1：

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]]
示例 2：

输入：nums = [0,1]
输出：[[0,1],[1,0]]
示例 3：

输入：nums = [1]
输出：[[1]]

提示：

1 <= nums.length <= 6
-10 <= nums[i] <= 10
nums 中的所有整数 互不相同

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

参考：

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

python

# 0046.全排列

class Solution:
    def permute(self, nums: [int]) -> [[int]]:
        res = []
        path = []
        used = [] # 存放已经使用过的数字
        def track(nums, used):
            if len(path) == len(nums):
                return res.append(path[:])
            for i in range(0, len(nums)):
                if nums[i] in used:
                    continue # used中已收录，跳过
                path.append(nums[i])
                used.append(nums[i])
                track(nums, used)
                used.pop()
                path.pop()

        track(nums, used)
        return res


# 不使用used记录使用的元素
class Solution:
    def permute(self, nums: [int]) -> [[int]]:
        res = []
        path = []
        def track(nums):
            if len(path) == len(nums):
                return res.append(path[:])
            for i in range(0, len(nums)):
                if nums[i] in path:
                    continue
                path.append(nums[i])
                track(nums)
                path.pop()

        track(nums)
        return res

golang

package backTrack

var res [][]int

func permute(nums []int) [][]int {
	res = [][]int{}
	track(nums, len(nums), []int{})
	return res
}

func track(nums []int, length int, path []int)  {
	if len(nums) == 0 {
		p := make([]int, len(path))
		copy(p, path)
		res = append(res, p)
	}
	for i:=0;i<length;i++ {
		cur := nums[i]
		path = append(path, cur)
		nums = append(nums[:i], nums[i+1:]...)
		track(nums, len(nums), path)
		nums = append(nums[:i], append([]int{}, nums[i:]...)...)
		path = path[:len(path)-1]
	}
}