LeetCode 040 Combination Sum II
4415 ワード
タイトル要求:Combination Sum II
Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
Each number in C may only be used once in the combination.
Note: All numbers (including target) will be positive integers. Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak). The solution set must not contain duplicate combinations.
For example, given candidate set
分析:
Combination Sumの中の要素は無限に使用できますが、Combination Sum IIは要素ごとに1回しか使用できません.
コードは次のとおりです.
Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
Each number in C may only be used once in the combination.
Note:
For example, given candidate set
10,1,2,7,6,1,5
and target 8
, A solution set is: [1, 7]
[1, 2, 5]
[2, 6]
[1, 1, 6]
分析:
Combination Sumの中の要素は無限に使用できますが、Combination Sum IIは要素ごとに1回しか使用できません.
コードは次のとおりです.
class Solution {
public:
vector<vector<int> > combinationSum2(vector<int> &candidates, int target) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
sort(candidates.begin(), candidates.end());
set<vector<int> > ans;
vector<int> record;
searchAns(ans, record, candidates, target, 0);
vector<vector<int> > temp;
for (set<vector<int> >::iterator it = ans.begin(); it != ans.end(); it++) {
temp.push_back(*it);
}
return temp;
}
private:
void searchAns(set<vector<int> > &ans, vector<int> &record, vector<int> &candidates, int target, int idx) {
if (target == 0) {
ans.insert(record);
return;
}
if (idx == candidates.size() || candidates[idx] > target) {
return;
}
for (int i = 1; i >= 0; i--) {
record.push_back(candidates[idx]);
}
for (int i = 1; i >= 0; i--) {
record.pop_back();
searchAns(ans, record, candidates, target - i * candidates[idx], idx + 1);
}
}
};