Permutations II

最新推荐文章于 2025-09-12 16:32:54 发布

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最新推荐文章于 2025-09-12 16:32:54 发布

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CC 4.0 BY-SA版权

文章标签： runtime

原文链接：http://www.cnblogs.com/amazingzoe/p/4873390.html

本文提供了一个C++实现的算法，用于生成包含重复元素的集合的所有唯一排列。通过递归调用和巧妙地避免重复交换元素来确保结果中没有重复的排列。

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].

Runtime: 28ms

 1 class Solution {
 2 public:
 3     vector<vector<int>> permuteUnique(vector<int>& nums) {
 4         vector<vector<int> > result;
 5         if(nums.empty()) return result;
 6         
 7         helper(result, nums, 0, nums.size() - 1);
 8         return result;
 9     }
10     
11     void helper(vector<vector<int> >& result, vector<int>& nums, int depth, int n){
12         if(depth == n){
13             result.push_back(nums);
14             return;
15         }
16         
17         for(int i = depth; i < nums.size(); i++){
18             if(noSwap(nums, depth, i)) continue;
19             
20             swap(nums[depth], nums[i]);
21             helper(result, nums, depth + 1, n);
22             swap(nums[depth], nums[i]);
23         }
24     }
25     
26     bool noSwap(vector<int>& nums, int depth, int i){
27         for(int j = depth; j < i; j++){
28             if(nums[j] == nums[i])
29                 return true;
30         }
31         return false;
32     }
33 };