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. 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] ]
方法：递归调用即可
class Solution { private: void recursive(int sum,vector<vector<int>>& res,vector<int>& combination,vector<int>& candidates, int target,int now){ if(sum==target){ res.push_back(combination); return ; } for(int i=now+1;i<candidates.size();++i){ if(i!=now+1&&candidates[i]==candidates[i-1]) continue; if(sum+candidates[i]>target) return; combination.push_back(candidates[i]); recursive(sum+candidates[i],res,combination,candidates,target,i); combination.pop_back(); } } public: vector<vector<int>> combinationSum2(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],res,combination,candidates,target,i); combination.pop_back(); } return res; } };转载于:https://www.cnblogs.com/GoFly/p/5751054.html
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