Leetcode Subsets II

Leetcode Subsets II，本问题与Subsets 类似，只是需要处理重复子集问题，这个问题，我们可以使用每次从长度为m的子集构成长度为m+1的子集时，只对等值的元素添加一次，这样就可以保证没有重复现象，处理代码与Subsets类似，如下：

#include<iostream>
#include<vector>
#include<algorithm>

using namespace std;

class Solution {
public:
vector<vector<int> > subsetsWithDup(vector<int>& nums) {
// Sort the nums for the left solution
sort(nums.begin(), nums.end());
int len = nums.size();
// Init the result vector re
vector<vector<int> > re;
re.push_back(vector<int>());
// The pos vector saves the last element postion of the re[i]
// in the nums.
vector<int> pos;
// The first element in re is empty, so the first value is -1
pos.push_back(-1);
// The start and end position of the length k element in re
int start = 0;
int end = 0;
while (re.back().size() != len) {
// The elements of re with length k can be formed by the elements
// of re with length k-1 append an element after its last element
for (int i = start; i <= end; i ++) {
for (int j = pos[i] + 1; j < len; j ++) {
// Every time construct the element with length k
// for element with length k-1 only include the elements
// in nums with equaling values once
if ((j == pos[i] + 1) || (j != 0 && nums[j] != nums[j - 1])) {
vector<int> temp(re[i]);
temp.push_back(nums[j]);
re.push_back(temp);
pos.push_back(j);
}
}
}
//Update the start and end position of elements with length k
start = end + 1;
end = re.size() - 1;
}
return re;
}
};

int main(int argc, char* argv[]) {
Solution so;
vector<int> test;
test.push_back(2);
test.push_back(1);
test.push_back(2);
test.push_back(3);
vector<vector<int> > re = so.subsetsWithDup(test);
for (int i = 0; i < re.size(); i ++) {
for (int j = 0; j < re[i].size();j ++) {
cout<<re[i][j]<<" ";
}
cout<<endl;
}
return 0;
}