Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidateswhere the candidate numbers sums to target.

Each number in candidatesmay only be used once in the combination.

Note:

All numbers (including target) will be positive integers.
The solution set must not contain duplicate combinations.

Input: candidates =[10,1,2,7,6,1,5], target =8,
A solution set is:
[
[1, 7],
[1, 2, 5],
[2, 6],
[1, 1, 6]
]
Example 2:

Input: candidates =[2,5,2,1,2], target =5,
A solution set is:
[
[1,2,2],
[5]
]

class Solution {
public:
    vector<pair<int,int>> freg;
    vector<vector<int
 
>> ans;
    vector<int> sequence;
    void dfs(int pos,int rest){
        if(rest == 0){
            ans.emplace_back(sequence);
            return;
        }
        if(pos == freg.size() || rest<freg[pos].first){
            return;//加上最後一個數後，超過了對應的target
        }
        dfs(pos+1,rest);

        
 
int most = min(rest/freg[pos].first,freg[pos].second);
        for(int i=1;i<=most;++i){
            sequence.emplace_back(freg[pos].first);
            dfs(pos+1,rest-i*freg[pos].first);
        }
        for(int i=1;i<=most;++i){
            sequence.pop_back();//恢復現場，為了便於後面的進一步遞迴
        }
    }
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(),candidates.end());
        for(int num:candidates){
            if(freg.empty() || num!=freg.back().first){
                freg.emplace_back(num,1);
            }
            else{
                ++freg.back().second;
            }
        }
        dfs(0,target);
        return ans;
    }
};

注意：1.回溯的方式，對於列表中的每個數都有兩種選擇，選擇或者不被選擇，類似一個二叉樹的方式進行遞迴。2.為了避免重複列舉，將所有一樣的數字放在一起，按照該數字可能出現的次數，逐個進行列舉判斷。