leetcode——Combination Sum

题目：

Given a set of candidate numbers (C) and a target number (T), find all unique combinations inC where the candidate numbers sums to T.

The same repeated number may be chosen fromC 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 target7,
A solution set is:

[  [7],  [2, 2, 3]]

解答：参考递归法求子集的方法 http://blog.csdn.net/zhaopengnju/article/details/51427484 稍加改变

class Solution {public:    void combine(vector<int>& candidates, int start, int sum, int target, vector<int>& tmp, vector<vector<int>>& result)    {        if (sum > target || start >= candidates.size())        {            return;        }        if (sum == target)        {            result.push_back(tmp);            return;        }        //不选当前元素        combine(candidates, start + 1, sum, target, tmp, result);        //选择当前元素        tmp.push_back(candidates[start]);        sum += candidates[start];        combine(candidates, start, sum, target, tmp, result);//由于同一个元素可以重复选取，所以依然从start递归        tmp.pop_back();    }    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {        vector<vector<int>> res;        if (candidates.empty())        {            return res;        }        //为了防止出现重复的集合，先对candidates进行排序        std::sort(candidates.begin(), candidates.end());        auto end = std::unique(candidates.begin(), candidates.end());        candidates.erase(end, candidates.end());        vector<int> tmp;        combine(candidates, 0, 0, target, tmp, res);        return res;    }};