问题描述：

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]
]

题意就是提供一个向量和目标值target，判断向量中能组成target的对并且输出，向量的元素可以使用多次。

解题思路：

深度优先遍历的方法，遍历函数为combine(vector<int>& candidates, int target, int index, vector<vector<int>>& result, vector<int> tmp)，参数的意思分别是原向量，目标值，当前遍历的索引，结果向量，当前判断的向量。

源码：

class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        vector<vector<int>> result;
        // result.clear();
        vector<int> tmp;
        combine(candidates, target, 0, result, tmp);
        return result;
    }
    void combine(vector<int>& array, int target, int index, vector<vector<int>>& result, vector<int> tmp){
        if(target<0)
            return;
        else if(target==0){
            result.push_back(tmp);
        }
        else{
            for(int i=index; i<array.size(); i++){
                tmp.push_back(array[i]);
                // cout<<target<<" "<<array[i]<<endl;
                combine(array, target-array[i], i, result, tmp);
                tmp.pop_back();
            }
        }
    }
};

改进的方法：

上面的做法时间复杂度很高，可以先给向量排序，然后再在深度搜索的时候多加一个条件target>=candidates[i]。从而大大的降低了时间复杂度。源码如下：

class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        vector<vector<int>> result;
        // result.clear();
        vector<int> tmp;
        combine(candidates, target, 0, result, tmp);
        return result;
    }
    void combine(vector<int>& candidates, int target, int index, vector<vector<int>>& result, vector<int> tmp){
        if(!target){
            result.push_back(tmp);
            return;
        }
        else{
            for(int i=index; i != candidates.size() && target>=candidates[i]; i++){
                tmp.push_back(candidates[i]);
                // cout<<target<<" "<<array[i]<<" "<<i<<endl;
                combine(candidates, target-candidates[i], i, result, tmp);
                tmp.pop_back();
            }
        }
    }
};