Given a collection of distinct numbers, return all possible permutations.

For example,
[1,2,3] have the following permutations:

[
  [1,2,3],
  [1,3,2],
  [2,1,3],
  [2,3,1],
  [3,1,2],
  [3,2,1]
]

思路：
采用逐一向中间集添加元素，并将当中间集元素个数等于 nums 长度的时候，将中间集添加到结果集中，并终止该层递归。

class Solution {
    public List<List<Integer>> permute(int[] nums) {  
        List<List<Integer>> res = new ArrayList<>();  
        dfs(res, new ArrayList<Integer>(), nums, new boolean[nums.length]);  
        return res;  
    }  
    private void dfs(List<List<Integer>> res, List<Integer> temp, int[] nums, boolean[] used) {  
        if(temp.size() == nums.length) {  
            res.add(new ArrayList<>(temp));  
            return;  
        }  
        for(int i = 0; i < nums.length; i++) {  
            if(!used[i]) {  
                used[i] = true;  
                temp.add(nums[i]);  
                dfs(res, temp, nums, used);  
                temp.remove(temp.size() - 1);  
                used[i] = false;  
            }  
        }  
    }  
}
class Solution {
    public List<List<Integer>> permute(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();
        permute(nums,0,nums.length-1, res);
        return res;
    }
    
    void permute(int[] nums,int l,int r, List<List<Integer>> res){
        if(l ==r){
            res.add(toList(nums));
        }else{
            for(int i =l;i<=r;i++){
                swap(nums,i,l);
                permute(nums,l+1,r, res);
                swap(nums,l,i);
            }
        }
    }
    
    List<Integer> toList(int[] src){
        List<Integer> dst = new ArrayList<>();
        for(Integer n: src){
            dst.add(n);
        }
        return dst;
    }
    
    void swap(int[]nums, int i ,int j){
        int temp = nums[i];
        nums[i]= nums[j];
        nums[j]= temp;
    }
}