Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
The same repeated number may be chosen from C unlimited number of times.
Note: All numbers (including target) will be positive integers. The solution set must not contain duplicate combinations. For example, given candidate set [2, 3, 6, 7] and target 7, A solution set is: [ [7], [2, 2, 3] ]
方法１:简单的递归调用。
class Solution { private: void recursive(int sum,vector<int> &combination,vector<int>& candidates,vector<vector<int>> &res,int target,int now){ if(sum>target) return; if(sum==target){ res.push_back(combination); return; } for(int i=now;i<candidates.size();i++){ if(i!=now&&candidates[i]==candidates[i-1]) continue; combination.push_back(candidates[i]); recursive(sum+candidates[i],combination,candidates,res,target,i); combination.pop_back(); } } public: vector<vector<int>> combinationSum(vector<int>& candidates, int target) { vector<vector<int>> res; vector<int> combination; sort(candidates.begin(),candidates.end()); for(int i=0;i<candidates.size();i++){ if(i!=0&&candidates[i]==candidates[i-1]) continue; combination.push_back(candidates[i]); recursive(candidates[i],combination,candidates,res,target,i); combination.pop_back(); } return res; } };转载于:https://www.cnblogs.com/GoFly/p/5751055.html
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