39. Combination Sum
Given asetof candidate numbers (C)(without duplicates)and a target number (T), find all unique combinations inCwhere the candidate numbers sums toT.
Thesamerepeated number may be chosen fromCunlimited number of times.
For example, given candidate set [2, 3, 6, 7] and target 7, A solution set is: [ [7], [2, 2, 3] ]
给一个数组C和一个目标值T，求C中数字的各种组合，使其和等于目标T，每个元素可以使用无限次。
用回溯法穷举就可以了，不过要限定只能向后选数字，不然会造成重复。具体在算法实现中，添加一个当前判定数字位置的pos变量就可以了。
1 class Solution { 2 public: 3 void helper(vector<int>& candidates, int target, int sum, int pos, vector<int> v, vector<vector<int>>& res) { 4 if (sum==target) { 5 res.push_back(v); 6 return; 7 } 8 for (int i=pos; i<candidates.size(); ++i) { 9 if (sum+candidates[i]<=target) { 10 v.push_back(candidates[i]); 11 sum += candidates[i]; 12 helper(candidates, target, sum, i, v, res); 13 sum -= candidates[i]; 14 v.pop_back(); 15 } 16 } 17 } 18 vector<vector<int>> combinationSum(vector<int>& candidates, int target) { 19 vector<vector<int>> res; 20 vector<int> v; 21 helper(candidates, target, 0, 0, v, res); 22 return res; 23 } 24 };优质内容筛选与推荐>>
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