[LeetCode]Permutations II

题目描述：

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

For example,
[1,1,2] have the following unique permutations:
[1,1,2], [1,2,1], and [2,1,1].

解题思路：

思路和上题目相似，但需要跳过重复的元素

 1 class Solution {
 2 public:
 3     vector<vector<int>> permuteUnique(vector<int>& nums) {
 4         sort(nums.begin(), nums.end());
 5         vector<vector<int>> result;
 6         vector<int> elem;
 7         vector<bool> visited(nums.size(), false);
 8         permute(nums, visited, result, elem);
 9         return result;
10     }
11 private:
12     void permute(const vector<int> &nums, vector<bool> &visited, vector<vector<int>> &result, vector<int> &elem) {
13         if (nums.size() == elem.size()) {
14             result.push_back(elem);
15             return;
16         }
17         
18         for (int i = 0; i < nums.size(); ++i) {
19             if (!visited[i]) {
20                 visited[i] = true;
21                 elem.push_back(nums[i]);
22                 permute(nums, visited, result, elem);
23                 elem.pop_back();
24                 visited[i] = false;
25                 while (i < nums.size() - 1 && nums[i] == nums[i + 1]) ++i; // 跳过重复的元素
26             }
27         }
28     }
29 };