LeetCode: Subsets
Given a set of distinct integers, 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 = [1,2,3], a solution is:
[ [3], [1], [2], [1,2,3], [1,3], [2,3], [1,2], [] ]
地址：https://oj.leetcode.com/problems/subsets/
算法：递归方法。先对数组排序，然后先构造前n-1个数的子集，则n个数的子集就包括两部分，1）前n-1个数子集；2）前n-1个数子集里各元素都加上第n个数。代码：
1 class Solution {
2 public:
3 vector<vector<int> > subsets(vector<int> &S) {
4 if(S.empty()) return vector<vector<int> >(1,vector<int>());
5 sort(S.begin(),S.end());
6 return subsetsCore(S,S.size());
7 }
8 vector<vector<int> > subsetsCore(vector<int> &S, int n){
9 vector<vector<int> > result;
10 if(n <= 0){
11 result.push_back(vector<int>());
12 return result;
13 }
14 vector<vector<int> > temp = subsetsCore(S, n-1);
15 result = temp;
16 for(int i = 0; i < temp.size(); ++i){
17 temp[i].push_back(S[n-1]);
18 result.push_back(temp[i]);
19 }
20 return result;
21 }
22 };
第二题：
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], [] ]
地址：https://oj.leetcode.com/problems/subsets-ii/
算法：数组里有重复的数，要求找出所有的子集。首先还是先对数组进行排序，然后构造只包含第一个元素以及空集这两个子集，然后从第二个数开始循环，若第二个数跟前面的数不一样，则把result里的子集个数加倍，其中多出来的子集为加上第二个数的
子集，若第二个数跟前面的数一样，则在加入新子集时要判断result中是否已经存在该子集。代码：
1 class Solution {
2 public:
3 vector<vector<int> > subsetsWithDup(vector<int> &S) {
4 vector<vector<int> > result;
5 if(S.empty()) result;
6 sort(S.begin(),S.end());
7 result.push_back(vector<int>(1,S[0]));
8 result.push_back(vector<int>());
9 for(int i = 1; i < S.size(); ++i){
10 int len = result.size();
11 if(S[i] != S[i-1]){
12 for(int j = 0; j < len; ++j){
13 vector<int> temp = result[j];
14 temp.push_back(S[i]);
15 result.push_back(temp);
16 }
17 }else{
18 for(int j = 0; j < len; ++j){
19 vector<int> temp = result[j];
20 temp.push_back(S[i]);
21 if(!isExist(result,temp,len)){
22 result.push_back(temp);
23 }
24 }
25 }
26 }
27 return result;
28 }
29 bool isExist(vector<vector<int> > &result, vector<int> &temp, int len){
30 int i = 0;
31 while(i < len && !isSameVector(result[i],temp)) ++i;
32 return i < len;
33 }
34 bool isSameVector(const vector<int> v1, const vector<int> v2){
35 if(v1.size() != v2.size())
36 return false;
37 int i = 0;
38 while(i < v1.size() && v1[i] == v2[i]) ++i;
39 return i == v1.size();
40 }
41 };