程序媛詹妮弗
终身学习

Given a collection of numbers that might contain duplicates, return all possible unique permutations.

Example:

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

 

题意:

打印全排列,注明了给定序列可含有重复元素

 

Solution1: Backtracking

 

code

 1 class Solution {
 2      public List<List<Integer>> permuteUnique(int[] nums) {
 3         List<List<Integer>> list = new ArrayList<>();
 4          List<Integer> path = new ArrayList<>();
 5         Arrays.sort(nums); // necessary!因为后面要查重
 6         dfs(list, path, nums, new boolean[nums.length]);
 7         return list;
 8 }
 9 
10     private void dfs(List<List<Integer>> list, List<Integer> path, int [] nums, boolean [] used){
11         if(path.size() == nums.length){
12             list.add(new ArrayList<>(path));
13             return;
14         }
15         for(int i = 0; i < nums.length; i++){
16             if(used[i] || i > 0 && nums[i] == nums[i-1] && !used[i - 1]) continue;
17             used[i] = true; //标记用过
18             path.add(nums[i]);
19             dfs(list, path, nums, used);
20             used[i] = false; //恢复default值
21             path.remove(path.size() - 1);
22         }
23     }
24 }

 

posted on 2019-04-15 02:43  程序媛詹妮弗  阅读(252)  评论(0编辑  收藏  举报