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 10,1,2,7,6,1,5 and target 8,
A solution set is:
[1, 7]
[1, 2, 5]
[2, 6]
[1, 1, 6]
先排序！
這個題，去重是關鍵；//[2,2,2,2] 4 怎麼快速去重呢？
先丟擲個很笨重的去重吧。40ms
解法1：

class Solution {
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(),candidates.end());
        vector<int> temp;
        vector<vector<int>> res;
        combinationSum(candidates,0,target,temp,res);
        return
 
 res;
    }
private:
    void combinationSum(vector<int>& candidates,int index,int target,vector<int> temp,vector<vector<int>> &res){
        if(index==candidates.size()||candidates[index]>target) return;//終止條件
        if(candidates[index]==target){
            temp.push_back(candidates[index]);
            if
 
(res.size()>0){//skip duplicates too slow!!!
                for(int i=0;i<res.size();i++){
                    if(equal(temp,res[i])) return;
                }
            }//不重複就加入！//但是這判斷重複的代價太大了吧？？？
            res.push_back(temp);//找到一組滿足條件的 
            return;
        }
         temp.push_back(candidates[index]);
         combinationSum(candidates,index+1,target-candidates[index],temp,res);
         temp.pop_back();
         combinationSum(candidates,index+1,target,temp,res);
    }

    bool equal(vector<int> &a,vector<int> &b){
        if(a.size()!=b.size()) return false;
        for(int i=0;i<a.size();i++){
            if(a[i]!=b[i]) return false;
        }
        return true;
    }
};

每次都是把整個res中的vector全部比較一遍，時間成本很高。
要想出在剛開始就去掉的方法！
解法2：把後面相同的元素跳過去。20ms

public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(),candidates.end());
        vector<int> temp;
        vector<vector<int>> res;
        combinationSum(candidates,0,target,temp,res);
        return res;
    }
private:
    void combinationSum(vector<int>& candidates,int index,int target,vector<int> temp,vector<vector<int>> &res){
      if(index==candidates.size()||candidates[index]>target) return;//終止條件
        for(int i=index;i<candidates.size();i++){
            if(i>index&&candidates[i]==candidates[i-1])
                  continue;
            temp.push_back(candidates[i]);
            if(candidates[i]==target){
               res.push_back(temp);//找到一組滿足條件的 
               return;
            }
            combinationSum(candidates,i+1,target-candidates[i],temp,res);
            temp.pop_back();
        }
    }