40. Combination Sum II
1.問題

2.想法
提問
時間複雜度與可用空間是否有限制?
member是否可以重複使用?
重複的排列是否視為同一個?
function header, parameter
test input
說明想法
這個問題有點類似39 Combination sum, 但允許member被重複使用
加入限制條件, 當candidates[i] == candidates[i - 1]時continue
測試計算複雜度
3.程式碼
class Solution {
public:
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
vector<vector<int>> res;
vector<int> record;
vector<bool> used(candidates.size(), false);
sort(candidates.begin(), candidates.end());
getCombinationSum(res, record, candidates, used, target, 0);
return res;
}
private:
void getCombinationSum(vector<vector<int>>& res, vector<int>& record, vector<int>& candidates, vector<bool>& used, int target, int start) {
if (target < 0) {
return;
}
if (target == 0) {
res.push_back(record);
return;
}
for (int i = start; i < candidates.size(); i++) {
if (i > start && candidates[i] == candidates[i - 1]){
continue;
}
record.push_back(candidates[i]);
getCombinationSum(res, record, candidates, used, target - candidates[i], i + 1);
record.pop_back();
}
}
};
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