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]
]

if(temp.contains(nums[i])) continue;temp里加过了的话，就不加了。这句话得有

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;
        
        //排序一下
        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
        if (temp.size() == nums.length)
            results.add(new ArrayList<>(temp));
        
        for (int i = 0; i < nums.length; i++) {
            // if ((i > start) && (nums[i] == nums[i - 1]))
            //     continue;
            
            if(temp.contains(nums[i])) continue;
            
            temp.add(nums[i]);
            dfs(nums, temp, i + 1, results);
            temp.remove(temp.size() - 1);
        }
 
    }
}