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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:
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<vector<int> > combinationSum2(vector<int> &candidates, int target) { results.clear(); vector<int> result; sort(candidates.begin(), candidates.end()); dfs(candidates, result,target, 0); return results; } void dfs(vector<int> &candidates, vector<int> &result, int target, int depth) { if(candidates[depth] > target || depth == candidates.size()) return; result.push_back(candidates[depth]); if(candidates[depth] == target) { results.push_back(result); result.pop_back(); return; } dfs(candidates,result, target-candidates[depth], depth+1); result.pop_back(); while(++depth<candidates.size() && candidates[depth] == candidates[depth-1]); //avoid repetition dfs(candidates,result, target, depth); } private: vector<vector<int> > results; };
40. Combination Sum II (Recursion)
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原文地址：http://www.cnblogs.com/qionglouyuyu/p/4855328.html
