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]
本题与上题类似，不过首先需要对数组进行排序，然后需要去除重复。时间：8ms。代码如下：
class Solution { public: void makeCombination2(vector<vector<int>>& ret, vector<int>& candidates, vector<int>& v, int target, int cur){ if (target == 0){ ret.push_back(v); return; } int rep = -1; for (size_t i = cur; i < candidates.size(); ++i){ if (rep == candidates[i]) continue; if (target >= candidates[i]){ rep = candidates[i]; v.push_back(candidates[i]); makeCombination2(ret, candidates, v, target - candidates[i], i + 1); v.pop_back(); } else return; } } vector<vector<int>> combinationSum2(vector<int>& candidates, int target) { vector<vector<int>> ret; sort(candidates.begin(), candidates.end()); if (candidates.empty() || target < candidates[0]) return ret; vector<int> v; makeCombination2(ret, candidates, v, target, 0); return ret; } };
转载于:https://www.cnblogs.com/Scorpio989/p/4581941.html
