[LeetCode]Combination Sum

最新推荐文章于 2019-05-21 16:08:00 发布

xshalk

最新推荐文章于 2019-05-21 16:08:00 发布

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本文介绍了一个经典的算法问题：给定一组候选数和目标数，找出所有可能的组合，使得组合中候选数之和等于目标数。同一数字可以无限次重复使用。文章提供了使用深度优先搜索（DFS）解决此问题的方法。

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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, … , a_k) must be in non-descending order. (ie, a₁ ≤ a₂ ≤ … ≤ a_k).
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]

先想到的是背包...不过要得到所有结果，而不是结果数量，就用了DFS

写了一下，又参考了一下，有两种思路，一种是统计各个数的需求个数，另一个是累加达到target 基本一致吧

class Solution {
private:
    vector<vector <int> > ret;
    vector <int> count;
public:
	void help(int current_index, int max_index, int target , vector<int> & candidates){

		if(target <0)
			return ;
        if(current_index == max_index)
		{
	    	if(target == 0)
			{
		        vector<int> tmp;
		    	for(int i = 0 ; i <  max_index; i++)
			    	for(int j = 0 ; j < count[i]; j++)
				    	tmp.push_back(candidates[i]);
		    	
		    	ret.push_back(tmp);
		    	return ;
	    	}
            return ;
         }
		for(int i = 0; i <= target/candidates[current_index] ; i++)
		{
			count[current_index] = i;
			help(current_index+1,max_index, target - i*candidates[current_index],candidates);
		}
	}
    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
     ret.clear();
    if(candidates.size()==0)
        return ret;
    sort (candidates.begin(),candidates.end()); 
    count.clear();
    count.resize(candidates.size());
	help(0,candidates.size(), target,candidates);
	return  ret;
    }
};