[LeetCode] Combination Sum

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