46. Permutations 全排列，无重复

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

Example:

Input: [1,2,3]
Output:
[
  [1,2,3],
  [1,3,2],
  [2,1,3],
  [2,3,1],
  [3,1,2],
  [3,2,1]
]

在其基础上扩展，所以是i + 1。要有个start index的参数

//跳过重复值，同一temp集合中不能出现两次
if(temp.contains(nums[i])) continue;

class Solution {
    public List<List<Integer>> permute(int[] nums) {
        //cc
        List<List<Integer>> results = new ArrayList<List<Integer>>();
        if (nums == null || nums.length == 0) 
            return results;
        
        Arrays.sort(nums);
        dfs(nums, new ArrayList<Integer>(), 0, results);
        
        return results;
    }
    
    public void dfs(int[] nums, List<Integer> temp, int start,
                    List<List<Integer>> results) {
        //exit 退出的条件，加在dfs里最外一层
       if (temp.size() == nums.length) {
           results.add(new ArrayList<>(temp));
       }
               
        for (int i = 0; i < nums.length; i++) {
            if (temp.contains(nums[i]))
                continue;
            
            temp.add(nums[i]);
            dfs(nums, temp, i + 1, results);
            temp.remove(temp.size() - 1);
        }
    }
}