Given a set of candidate numbers (C) 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:
For example, given candidate set 2,3,6,7 and
 target 7, 
A solution set is: 
[7] 
[2, 2, 3] 
class Solution {
public:
    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
        vector<vector<int>> result;
        vector<int> path;
        result.clear();
        path.clear();
        sort(candidates.begin(),candidates.end());
        GetPath(candidates,target,0,0,path,result);
        return result;
        
    }
    void GetPath(vector<int> &candidates, int target,int depth,int sum,vector<int> &path,vector<vector<int>> &result)
    {
        if(sum > target)
        {
            return;
        }
        if(sum == target)
        {
            result.push_back(path);
            return;
        }
        int i;
        for(i = depth; i < candidates.size(); i++)
        {
            path.push_back(candidates[i]);
            GetPath(candidates,target,i,sum+candidates[i],path,result);
            path.pop_back();
        }
        return;
    }
};主要是找数组中能够加起来等于一个特定的数字
用的是DFS的算法,感觉DFS算法还是比较有特点的,就是在搜索的过程中,只能满足部分数据是进行满足的,不知道后面的情况
通过一个深度信息,和一个比如sum值进行不断的向下搜索,如果搜到了最后满足了一定的条件,就表示这个路径是可以考虑的。
LeetCode—**Combination Sum 利用DFS算法
原文:http://blog.csdn.net/xietingcandice/article/details/44516331