39. Combination Sum

Problem:

Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

The same repeated number may be chosen from candidates unlimited number of times.

Note:

• All numbers (including target) will be positive integers.
• The solution set must not contain duplicate combinations.

Example 1:

Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
  [7],
  [2,2,3]
]

Example 2:

Input: candidates = [2,3,5], target = 8,
A solution set is:
[
  [2,2,2,2],
  [2,3,3],
  [3,5]
]

思路：

采用回溯法(BackTracking)。

Solution (C++):

public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        vector<vector<int>> res;
        vector<int> combination;
        combinationSum(candidates, target, res, combination, 0);
        
        return res;
    }
private:
    void combinationSum(vector<int> &candidates, int target, vector<vector<int>> &res, vector<int> &combination, int begin) {
        if (target == 0) {
            res.push_back(combination);
            return ;
        }
        for (int i = begin; i < candidates.size() && target >= candidates[i]; ++i) {
            combination.push_back(candidates[i]);
            combinationSum(candidates, target - candidates[i], res, combination, i);    //不是解则无变化，pop掉尾部的数
            combination.pop_back(candidates[i]);
        }
    }

性能：

Runtime: 12 ms  Memory Usage: 9.5 MB