[LeetCode] Subsets II 解题报告

最新推荐文章于 2019-07-27 20:45:00 发布

原创最新推荐文章于 2019-07-27 20:45:00 发布 · 179 阅读

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本文介绍了一种解决含重复元素集合的所有可能子集问题的算法。通过递归生成子集，并在过程中去除重复项，确保结果中不含相同的子集。文章提供了C++实现代码，并与不含重复元素的情况进行了对比。

Given a collection of integers that might contain duplicates, S, return all possible subsets.

Note:

Elements in a subset must be in non-descending order.
The solution set must not contain duplicate subsets.

For example,
If S = [1,2,2], a solution is:

[
  [2],
  [1],
  [1,2,2],
  [2,2],
  [1,2],
  []
]

» Solve this problem

[解题思路]
跟Subset I思路一样，http://codingtmd.azurewebsites.net/leetcode-subsets-%e8%a7%a3%e9%a2%98%e6%8a%a5%e5%91%8a/。区别只在于去重。
代码的区别只有Line 24， 25。对于当前字符，如果下一个字符与之相等，则过滤掉。

[Code]

1:    vector<vector<int> > subsetsWithDup(vector<int> &S) {  
2:      // Start typing your C/C++ solution below    
3:      // DO NOT write int main() function  
4:      vector<vector<int> > result;  
5:      vector<int> output;   
6:      if(S.size() ==0) return result;  
7:      result.push_back(output);  
8:      sort(S.begin(), S.end());  
9:      generateSub(S, 0, result, output);  
10:    }  
11:    void generateSub(  
12:      vector<int> &s,   
13:      int step,   
14:      vector<vector<int> > &result,  
15:      vector<int>& output)  
16:    {         
17:      for(int i = step;i<s.size(); i++ )  
18:      {  
19:        output.push_back(s[i]);  
20:        result.push_back(output);  
21:        if(i< s.size()-1)  
22:          generateSub(s, i+1, result, output);  
23:        output.pop_back();  
24:        while(i<s.size()-1 && s[i] == s[i+1])  
25:          i++;  
26:      }      
27:    }