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[LeetCode] Combination Sum II （递归）
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 （递归）
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