标签：des style blog os io strong for div cti
Given a set of distinct integers, S, return all possible subsets.
Note:
For example, If S = [1,2,3], a solution is:
[ [3], [1], [2], [1,2,3], [1,3], [2,3], [1,2], [] ]
思想： 顺序读，取前面的每个子集，把该位置数放后面作为新的子集。
class Solution {
public:
vector<vector<int> > subsets(vector<int> &S) {
sort(S.begin(), S.end());
vector<vector<int> > vec(1);
for(size_t id = 0; id < S.size(); ++id) {
int n = vec.size();
while(n-- > 0) {
vec.push_back(vec[n]);
vec.back().push_back(S[id]);
}
}
return vec;
}
};
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], [] ]
思想： 排序后，按照 1 的方法。但是若前面的数字与本数字相同，则只读取含有前面数字的每个子集，把自身放在后面作为一个新的子集。
class Solution {
public:
vector<vector<int> > subsetsWithDup(vector<int> &S) {
sort(S.begin(), S.end());
vector<vector<int> > vec(1);
size_t prePos, endTag;
prePos = endTag = 0;
for(size_t id = 0; id < S.size(); ++id) {
if(id > 0 && S[id] != S[id-1]) endTag = 0;
else endTag = prePos;
size_t n = vec.size();
prePos = n;
while(n > endTag) {
--n;
vec.push_back(vec[n]);
vec.back().push_back(S[id]);
}
}
return vec;
}
};
标签：des style blog os io strong for div cti
原文地址：http://www.cnblogs.com/liyangguang1988/p/3940198.html
