Leetcode: Combination Sum II

Combination Sum II

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Total Accepted: 54743 Total Submissions: 208167 Difficulty: Medium

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] 

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 Array Backtracking

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o Judge: sum > target?

o Judge: sum==target?

o Process: sum+=source[i];

                  BT;

                  

///////////////////////////////////////

//codes

class Solution {

public:

   void backTrac(vector<int>& cand, int i, int temSum, int target, map<vector<int>,int>& mapT, vector<int>& vec){

      

       if(temSum==target){mapT[vec]++;return ;}

       else if(temSum>target){return ;}

       else{

           for (;i<cand.size();i++){

              temSum+=cand[i];

              vec.push_back(cand[i]);

              if (temSum>target) i = cand.size();//remove the unnecessary implementations

              backTrac(cand, i+1, temSum, target, mapT, vec);

              vec.pop_back();

              temSum-=cand[i];

           }

           return ;

       }

   }

    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {

        //if(candidates.size()==0)

        //sort

        sort(candidates.begin(), candidates.end());

        map<vector<int>,int> mapT;

        vector<int> vec;

        int temSum=0;

        backTrac(candidates, 0, temSum, target, mapT, vec);

        vector<vector<int>> res;

        map<vector<int>,int>::iterator it=mapT.begin();

        for (int i=0;i<mapT.size();i++, it++){

            res.push_back(it->first);

        }

        return res;

    }

};