题目链接：
https://leetcode-cn.com/problems/combination-sum-ii/submissions/
难度：中等
40. 组合总和 II
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。
说明：
所有数字（包括目标数）都是正整数。
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,
所求解集为:
[
[1, 7],
[1, 2, 5],
[2, 6],
[1, 1, 6]
]
示例 2:
输入: candidates = [2,5,2,1,2], target = 5,
所求解集为:
[
[1,2,2],
[5]
]
(The freshman registration status is not good...)
This question won’t know how to remove duplicates...
In the recursive process, if the number x appears y times, then deal with them all at once during the recursion, that is, call to select 0, respectively 1,...,y recursive function of x.
(It’s too bad to have time to rewrite according to the idea of the problem solution ...) Problem solution
class Solution {
public:
vector<vector<int>> ans;
vector<pair<int,int>> freq;
vector<int> sequence;
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;
}
void dfs(int pos,int rest){
if(rest==0){
ans.push_back(sequence);
}
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();
}
}
};
(That’s it if the state is not good)