Combination Sum II

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]

class Solution {
public:
    vector<vector<int> > ans;
    vector<vector<int> > combinationSum2(vector<int> &num, int target){
        /**
            10,1,2,7,6,1,5 and target 8,
            [1, 7] 
            [1, 2, 5] 
            [2, 6] 
            [1, 1, 6]
            
            Each number in C may only be used once in the combination.
        */
        vector<int> path;
        ans.clear();
        
        if (num.size() == 0){
            return ans;
        }
        //sort
        vector<pair<int, int> > num_counter;
        sort(num.begin(),num.end());
        
        int counter = 1;
        for(int i = 1; i < num.size(); i++){
            if (num[i] == num[i -1]){
                counter++;
            }else{
                num_counter.push_back(make_pair(num[i-1],counter));
                counter = 1;
            }
        }
        num_counter.push_back(make_pair(num[num.size() -1],counter));
        
        dfs(num_counter,path,target);
        return ans;
    }
    void dfs(vector<pair<int,int> > & candidates, vector<int> & path,int target){
            if (target == 0){
                ans.push_back(path);
                return;
            }
            if (target < 0){
                return;
            }
            int j = 0;
            for(;j < candidates.size();j++){
                if (path.empty()){
                    break;
                }
                if (candidates[j].first == path[path.size() -1]){
                    int c = 0;
                    for(int m = 0; m < path.size(); m++){
                        if (path[m] == candidates[j].first){
                            c++;
                        }
                    }
                    if (c < candidates[j].second){
                        break;
                    }
                }else if (candidates[j].first > path[path.size() -1]){
                    break;
                }
            }
            for(int i = j; i < candidates.size() && target - candidates[i].first >= 0; i++){
                //candidates 事先排好序
                path.resize(path.size() + 1);
                path[path.size() -1] = candidates[i].first;
                dfs(candidates,path,target - candidates[i].first);
                path.resize(path.size() - 1);
        }
    }
};

using namespace std;
int main(int argc, char *argv[]) {
    int a[] = {1,2,1,5,1,4,6,7};
    vector<int> v(a,a+8);
    vector<vector<int> > ans;
    Solution sol;
    ans = sol.combinationSum2(v,7);
    for(int i = 0; i < ans.size(); i++){
        for(int j = 0; j < ans[i].size(); j++){
            cout<<" " << ans[i][j];
        }
        cout << endl;
    }
}