题目
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],
[]
]
思路
和78.Subsets的唯一区别就是添加了两行去重的代码。
代码
/**------------------------------------
* 日期:2015-03-01
* 作者:SJF0115
* 题目: 90.Subsets II
* 网址:https://oj.leetcode.com/problems/subsets-ii/
* 结果:AC
* 来源:LeetCode
* 博客:
---------------------------------------**/
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
class Solution {
public:
vector<vector<int> > subsetsWithDup(vector<int> &S) {
int size = S.size();
vector<vector<int> > result;
vector<int> path;
// 排序
sort(S.begin(),S.end());
// 空集
result.push_back(path);
// 其他子集
for(int i = 1;i <= size;++i){
DFS(S,size,i,0,path,result);
}//for
return result;
}
private:
// s源数据集 n源数据个数 k子集长度 index为第index个元素 path路径 result最终结果
void DFS(vector<int> &s,int n,int k,int index,vector<int> &path,vector<vector<int> > &result){
// 一个子集
if(path.size() == k){
result.push_back(path);
return;
}//if
for(int i = index;i < n;++i){
// 去重
if(i != index && s[i] == s[i-1]){
continue;
}//if
path.push_back(s[i]);
DFS(s,n,k,i+1,path,result);
path.pop_back();
}//for
}
};
int main(){
Solution s;
vector<int> num = {1,2,2};
vector<vector<int> > result = s.subsetsWithDup(num);
// 输出
for(int i = 0;i < result.size();++i){
for(int j = 0;j < result[i].size();++j){
cout<<result[i][j]<<" ";
}//for
cout<<endl;
}//for
return 0;
}
运行时间
原文地址:http://blog.csdn.net/sunnyyoona/article/details/44007093