Given a set of candidate numbers (C) (without duplicates) 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] ]
使用回溯法求解。
主要是对于辅助函数helper的构建
void helper(vector<vector<int>>& res, vector<int>& tmp, vector<int>& candidates, int target, int idx);
idx为遍历的头索引。
helper(res, tmp, candidates[i], target - candidates[i], i);
每次将target更新为选择池中的剩余目标值。i不变是因为组成target的值允许重复。
class Solution { public: vector<vector<int>> combinationSum(vector<int>& candidates, int target) { vector<vector<int>> res; vector<int> tmp; int idx = 0; helper(res, tmp, candidates, target, idx); return res; } void helper(vector<vector<int>>& res, vector<int>& tmp, vector<int>& candidates, int target, int idx) { if (target < 0) { return; } else if (target == 0) { res.push_back(tmp); } else { for (int i = idx; i < candidates.size(); i++) { tmp.push_back(candidates[i]); helper(res, tmp, candidates, target - candidates[i], i); tmp.pop_back(); } } } }; // 16 ms
