标签：not element blank 返回 ble pre str elements .com
Given a collection of integers that might contain duplicates, S, return all possible subsets.
Note:
For example,
If S =[1,2,2], a solution is:
[ [2], [1], [1,2,2], [2,2], [1,2], [] ]
题意：可能有重复的元素，返回所有子集。
思路：这题是subsets的扩展。大致的思路也是相同的，先是对给定整数进行排序，这样就可以保证所求的子集都是非降序的；在循环或递归的过程中跳过重复的。DFS法的代码如下：
1 class Solution { 2 public: 3 vector<vector<int> > subsets(vector<int> &S) 4 { 5 vector<vector<int>> res; 6 vector<int> midArray; 7 sort(S.begin(),S.end()); 8 getSubsets(S,0,midArray,res); 9 10 return res; 11 } 12 13 void getSubsets(vector<int> &S,int beg,vector<int> &midArray,vector<vector<int>> &res) 14 { 15 res.push_back(midArray); 16 for(int i=beg;i<S.size();++i) 17 { 18 midArray.push_back(S[i]); 19 getSubsets(S,i+1,midArray,res); 20 midArray.pop_back(); 21 22 while(S[i]==S[i+1]) //跳过重复的元素 23 i++; 24 } 25 } 26 };
迭代法：代码来源Grandyang的博客
1 class Solution { 2 public: 3 vector<vector<int> > subsets(vector<int> &S) 4 { 5 vector<vector<int>> res(1); 6 if(S.size()==0) return res; 7 8 sort(S.begin(),S.end()); 9 int size=res.size(); 10 if(S.size()==0) return res; 11 int last=S[0]; 12 13 for(int i=0;i<S.size();++i) 14 { 15 if(last !=S[i]) 16 { 17 last=S[i]; 18 size=res.size(); 19 } 20 21 int newSize=res.size(); 22 for(int j=newSize-size;j<newSize;++j) 23 { 24 res.push_back(res[j]); 25 res.back().push_back(S[i]); 26 } 27 } 28 return res; 29 } 30 };
[Leetcode] subsets ii 求数组所有的子集
标签：not element blank 返回 ble pre str elements .com
原文地址：http://www.cnblogs.com/love-yh/p/7126813.html
