【Leetcode】【Medium】Subsets II
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],
[]
]
解题:
先对S进行排序,然后继续按照Subsets I的思路,
由于出现了重复元素,需要分别考虑:
当出现的数字和上一个数字不同,保持原返回队列不变,对新出现的数字,分别插入已有数组的后面,形成新数组,加入返回值队列中;此时记录形成的新数组个数X;
当出现的数字和上一个数字相同,保持原返回队列不变,从原返回队列的后方遍历X个已有数组,对这X个数组插入这个相同的数字,形成新数组,加入返回队列中;X值不变;
代码:
class Solution {
public:
vector<vector<int> > subsetsWithDup(vector<int> &S) {
sort(S.begin(), S.end());
vector<vector<int> > ret;
ret.push_back(vector<int> ());
int duplicate_record = ; for (int i = ; i < S.size(); ++i) {
int pre_size = ret.size(); if (i== || S[i] != S[i-]) {
for (int j = ; j < pre_size; ++j) {
vector<int> new_item(ret[j]);
new_item.push_back(S[i]);
ret.push_back(new_item);
}
duplicate_record = ret.size() / ; } else if (S[i] == S[i-]) {
for (int j = pre_size - ; j >= pre_size - duplicate_record; --j) {
vector<int> new_item(ret[j]);
new_item.push_back(S[i]);
ret.push_back(new_item);
}
} } return ret;
}
};
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