Combination Sum II

Q: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, � , a_k) must be in non-descending order. (ie, a₁ ? a₂ ? � ? a_k).
• 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] 

A: 考虑有重复元素的情况[1,1] 2 和[1] ,1。引入bvisited数组

    vector<vector<int> > combinationSum2(vector<int> &num, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<vector<int> > result;
        vector<int> path;
        vector<bool> bvisited(num.size(),false);
        
        sort(num.begin(),num.end());
        
        combinationsum_aux(num,target,0,0,bvisited,path,result);
        return result;
        
    }
    
    bool combinationsum_aux(vector<int>& num,int target,int curPos,int curSum,vector<bool>& bvisited,vector<int> &path,vector<vector<int> >& result)
    {
        if(curSum==target)
        {
            result.push_back(path);
            return true;
        }else if(curSum>target)
            return false;
                
        for(int i=curPos;i<num.size();i++)
        {
            if(i>0&&num[i]==num[i-1]&&!bvisited[i-1])  //注意这里
                continue;
            path.push_back(num[i]);
            bvisited[i] = true;
            if(!combinationsum_aux(num,target,i+1,curSum+num[i],bvisited,path,result))
            {
                path.pop_back(); 
                bvisited[i] = false;
                break;              //ATT： return ture not false。此时表示这一层访问结束。
            }
            path.pop_back();
            bvisited[i] = false;
        }
        return true;
    }