LeetCode | 39. Combination Sum

组合求和算法解析

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本文介绍了一种使用回溯法解决组合求和问题的算法。该算法从候选数字集合中找出所有可能的组合，这些组合的元素之和等于目标值。文章详细解释了递归过程，并提供了完整的C++代码实现。

Given a set of candidate numbers (C) (without duplicates) 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.
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]
]

解题思路：回溯法。从前向后枚举，对于当前位置的元素，选择将它放入集合中或者跳过这个元素考察下一个元素，很好实现。

class Solution {
public:
    void f(vector<int> &candidates,int target,vector<vector<int> > &res,vector<int> &tmp,int _index)
    {
        if(target < 0)
            return;
        if(target == 0)
        {
            res.push_back(tmp);
            return;
        }
        for(int i=_index;i<candidates.size();i++)
        {
            tmp.push_back(candidates[i]);
            f(candidates,target-candidates[i],res,tmp,i);//当前元素放入集合
            tmp.pop_back();
        }
    }

    vector<vector<int> > combinationSum(vector<int>& candidates, int target)
    {
        sort(candidates.begin(),candidates.end());
        vector<vector<int> > res;
        vector<int> tmp;
        f(candidates,target,res,tmp,0);
        return res;
    }
};