leetcode 47. Permutations II

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]
]

思路

 

 1 class Solution {
 2 public:
 3     vector<vector<int>> permuteUnique(vector<int>& nums) {
 4         int len = nums.size();
 5         sort(nums.begin(), nums.end());
 6         vector<int> v;
 7         vector<vector<int> > res;
 8         vector<bool> visited(len, false);
 9         permuteUnique(nums, v, res, visited, 0, len);
10         return res;
11     }
12     //visited用于判断是否访问过， i表示v的长度， len表示nums的长度
13     void permuteUnique(vector<int> &nums, vector<int> &v, vector<vector<int> > &res, vector<bool> &visited, int i, int len) {
14         if (i == len) {
15             res.push_back(v);
16             return ;
17         }
18         for (int k = 0; k < len; k++) {
19             if (k > 0 && nums[k] == nums[k - 1] && visited[k - 1]) {
20                 continue;
21             }
22             if (visited[k] == false) {
23                 visited[k] = true;
24                 v.push_back(nums[k]);
25                 permuteUnique(nums, v, res, visited, i + 1, len);
26                 v.pop_back();
27                 visited[k] = false;
28             }
29         }
30     }
31 
32 };

 

posted @ 2019-09-19 16:04  琴影  阅读(157)  评论(0)  编辑  收藏  举报