46. Permutations-全排列

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

1.回溯：
Python:

class Solution:
    def permute(self, nums: List[int]) -> List[List[int]]:
        res = []
        flag = [True for i in range(len(nums))]
        def dfs(nums, temp):
            if len(temp) == len(nums):
                res.append(deepcopy(temp))
            for i in range(len(nums)):
                if flag[i]:
                    temp.append(nums[i])
                    flag[i] = False
                    dfs(nums, temp)
                    temp.pop()
                    flag[i] = True
        dfs(nums, [])
        return res

Java:

class Solution {
    public List<List<Integer>> permute(int[] nums) {
        List<List<Integer>> arr = new ArrayList<List<Integer>>();
        List<Integer> brr = new ArrayList<Integer>();
        int[] flag = new int[nums.length];
        recur(arr, nums, flag, brr);
        return arr;
    }

    public void recur(List<List<Integer>> a, int[] nums, int[] flag, List<Integer> b)
    {
        if(b.size() == nums.length)
        {
            a.add(new ArrayList<Integer>(b));
        }
        for(int i = 0; i < nums.length; i++)
        {
            if(flag[i] == 0)
            {
                flag[i] = 1;
                b.add(nums[i]);
                recur(a, nums, flag, b);
                b.remove(b.size()-1);
                flag[i] = 0;
            }
        }
    }

}

2.递归
Python:

class Solution:
    def permute(self, nums: List[int]) -> List[List[int]]:
        res = []
        def recur(nums, p):
            if p == len(nums):
                res.append(deepcopy(nums))
            for i in range(p, len(nums)):
                nums[i], nums[p] = nums[p], nums[i]
                recur(nums, p+1)
                nums[i], nums[p] = nums[p], nums[i]
        recur(nums, 0)
        return res

Java:

class Solution {
    public List<List<Integer>> permute(int[] nums) {
        List<List<Integer>> arr = new ArrayList<List<Integer>>();
        recur(arr, nums, 0);
        return arr;
    }

    public void recur(List<List<Integer>> a, int[] nums, int index)
    {
        if(index == nums.length)
        {   
            List<Integer> temp = Arrays.asList(Arrays.stream(nums.clone()).boxed().toArray(Integer[]::new));
            a.add(temp);
        }

        for(int i = index; i < nums.length; i++)
        {
            int temp1 = nums[i];
            nums[i] = nums[index];
            nums[index] = temp1;
            recur(a, nums, index+1);
            int temp2 = nums[i];
            nums[i] = nums[index];
            nums[index] = temp2;
        }
    }
}