【重点】【回溯】46.全排列

题目
跟另外一个题目很像：字符串的排列

Python

class Solution:
    def permute(self, nums: List[int]) -> List[List[int]]:
        res = list()
        self.dfs(nums, 0, res)
        return res
    
    def dfs(self, nums, start, res):
        if start == len(nums) - 1:
            res.append(nums.copy()) # list自带copy方法
            return 
        
        for i in range(start, len(nums), 1):
            nums[start], nums[i] = nums[i], nums[start]
            self.dfs(nums, start + 1, res)
            nums[start], nums[i] = nums[i], nums[start]

Java

法1：DFS，最佳解法

class Solution {
    public List<List<Integer>> permute(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();
        if (nums.length == 0) {
            return res;
        }
        dfs(nums, 0, res);

        return res;
    }

    public void dfs(int[] nums, int curInx, List<List<Integer>> res) {
        if (curInx == nums.length - 1) {
            List<Integer> tmp = new ArrayList<>();
            for (int i = 0; i < nums.length; ++i) {
                tmp.add(nums[i]);
            }
            res.add(tmp);
            return;
        }
        
        for (int i = curInx; i < nums.length; ++i) {
            swap(nums, curInx, i);
            dfs(nums, curInx + 1, res);
            swap(nums, curInx, i);
        }
    }

    public void swap(int[] array, int i, int j) {
        int tmp = array[i];
        array[i] = array[j];
        array[j] = tmp;
    }
}