给定一个数组 candidates 和一个目标数 target ,找出 candidates 中所有可以使数字和为 target 的组合。
candidates 中的每个数字在每个组合中只能使用一次。
注意:解集不能包含重复的组合。
示例 1:
输入: candidates = [10,1,2,7,6,1,5], target = 8,
输出:
[
[1,1,6],
[1,2,5],
[1,7],
[2,6]
]
示例 2:
输入: candidates = [2,5,2,1,2], target = 5,
输出:
[
[1,2,2],
[5]
]
提示:
- 1 <= candidates.length <= 100
- 1 <= candidates[i] <= 50
- 1 <= target <= 30
class Solution {
private:
vector<pair<int, int>> freq;
vector<vector<int>> ans;
vector<int> sequence;
public:
void dfs(int pos, int rest) {
if (rest == 0) {
ans.push_back(sequence);
return;
}
if (pos == freq.size() || rest < freq[pos].first) {
return;
}
dfs(pos + 1, rest);
int most = min(rest / freq[pos].first, freq[pos].second);
for (int i = 1; i <= most; ++i) {
sequence.push_back(freq[pos].first);
dfs(pos + 1, rest - i * freq[pos].first);
}
for (int i = 1; i <= most; ++i) {
sequence.pop_back();
}
}
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
sort(candidates.begin(), candidates.end());
for (int num: candidates) {
if (freq.empty() || num != freq.back().first) {
freq.emplace_back(num, 1);
} else {
++freq.back().second;
}
}
dfs(0, target);
return ans;
}
};