[LeetCode]Combination Sum II

最新推荐文章于 2019-05-10 10:55:45 发布

xshalk

最新推荐文章于 2019-05-10 10:55:45 发布

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本文链接：https://blog.youkuaiyun.com/xshalk/article/details/8162992

本文介绍了一种寻找候选数字集合中所有唯一组合的算法，这些组合的数字之和等于目标数字。该算法确保了组合中每个数字仅使用一次，并且遵循非递减顺序的要求。通过示例展示了如何为给定的数字集找到所有可能的有效组合。

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

Each number in C may only be used once in the combination.

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 10,1,2,7,6,1,5 and target 8,
A solution set is:
[1, 7]
[1, 2, 5]
[2, 6]
[1, 1, 6]

有点像01背包，稍微改动了一下，先预处理去重，统计每个元素的数，深度遍历的时候约束条件添加不能超过Num[i]

class Solution {
private:
    vector<vector <int> > ret;
    vector <int> count;
    vector <int> num;
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 <= num[current_index] && 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> > combinationSum2(vector<int> &candidates, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
     ret.clear();
	 vector <int> can;
	 int pre = 0;
    if(candidates.size()==0)
        return ret;
    sort (candidates.begin(),candidates.end()); 

    num.clear();
    num.resize(candidates.size());
    int j = -1;
	for(int i = 0 ; i < candidates.size(); i++)
	{
		if(candidates[i]==pre)
		{
			num[j]++;
		}else{
			j++;
			can.push_back(candidates[i]);
            pre =candidates[i];
			num[j] = 1;
		}
	}
    count.clear();
    count.resize(candidates.size());
	help(0,can.size(), target,can);
	return  ret;
    }
};