[LeetCode] Combination Sum II （遞迴）

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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]

class Solution {public:    vector<int> num;    int target;    vector<vector<int> > result;    vector<vector<int> > combinationSum2(vector<int> &num, int target) {        this->num = num;        this->target = target;                if(num.size()==0){            vector<int> tmp;            result.push_back(tmp);            return result;        }                sort(this->num.begin(),this->num.end());        for(int i=0;i<num.size();i++){           vector<int> tmp(1,this->num[i]);           recursion(i+1,tmp,this->num[i]);                }                return result;    }private:    void recursion(int start,vector<int> tmp,int sum){        if(sum == target && find(result.begin(),result.end(),tmp)==result.end()){            result.push_back(tmp);            return;        }                            vector<int> tmp0(tmp);        int sum0 = sum;        for(int i=start;i<num.size();i++){            tmp.push_back(num[i]);            sum += num[i];            if(sum<target)               recursion(i+1,tmp,sum);            else if(sum == target){                if(find(result.begin(),result.end(),tmp)==result.end())                    result.push_back(tmp);            }                            else                break;            tmp = tmp0;            sum = sum0;        }        }//end func};

 

[LeetCode] Combination Sum II （遞迴）