小小程序媛  
得之坦然,失之淡然,顺其自然,争其必然

题目

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:

All numbers (including target) will be positive integers.

Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).

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]

分析

用递归的思想实现~

AC代码

class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        if (candidates.empty() || target < 0)
            return vector<vector<int> >();

        ret.clear();
        //将给定序列排序
        sort(candidates.begin(), candidates.end());

        vector<int> tmp;
        combination(candidates, 0, tmp, target);
        return ret;
    }

    //递归实现
    void combination(vector<int> &candidates, int idx, vector<int> &tmp, int target)
    {
        if (target == 0)
        {
            ret.push_back(tmp);
            return;
        }
        else{
            int len = candidates.size();
            for (int i = idx; i < len; i++)
            {
                if (target >= candidates[i])
                {
                    tmp.push_back(candidates[i]);
                    combination(candidates, i, tmp, target - candidates[i]);
                    tmp.pop_back();
                }//if
            }//for
        }//else
    }

private:
    vector<vector<int> > ret;
};

GitHub测试程序源码

posted on 2015-10-04 17:03  Coding菌  阅读(121)  评论(0编辑  收藏  举报