LeetCode-4Sum

寻找数组中和为目标值的四元组

最新推荐文章于 2022-02-10 12:14:07 发布

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最新推荐文章于 2022-02-10 12:14:07 发布

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原文链接：http://www.cnblogs.com/superzrx/p/3330731.html

本文提供了一种复杂度为O(n^2)的算法来解决寻找数组中四个元素之和等于目标值的问题，确保这些元素在数组中按非递减顺序排列且不包含重复的四元组。

Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note:

Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
The solution set must not contain duplicate quadruplets.

For example, given array S = {1 0 -1 0 -2 2}, and target = 0.

    A solution set is:
    (-1,  0, 0, 1)
    (-2, -1, 1, 2)
    (-2,  0, 0, 2)
复杂度：O(n^2)
用两个for循环以及预处理过的数对

class Solution{
    public:
	vector<vector<int> > fourSum(vector<int> &num, int target) {

		unordered_map<int,int> m;
		//O(n) initialize
		for(int i=0;i<num.size();i++){
			if(m.find(num[i])==m.end()){
				m[num[i]]=1;
			}
			else{
				m[num[i]]++;
			}
		}
		//O(n^2) initialize
		unordered_map<int, set<vector<int> > > m2;
		vector<int> p;
		p.resize(2);
		for(int i=0;i<num.size();i++){
			for(int j=i+1;j<num.size();j++){
				int b,s;
				if(num[i]>num[j]){
					b=num[i];
					s=num[j];
				}
				else{
					b=num[j];
					s=num[i];
				}
				p[0]=s;
				p[1]=b;
				m2[b+s].insert(p);
			}
		}
		vector<int> one;
		one.resize(4);
		set<vector<int> >::iterator it;

		set<vector<int> >res;
		vector<vector<int> > ret;
		for( int i=0;i<num.size();i++)
		{
			for(int j=i+1;j<num.size();j++){
				int tt=target-num[i]-num[j];
				if(m2.find(tt)!=m2.end()){
					set<vector<int> >& ss=m2[tt];
					for(it=ss.begin();it!=ss.end();it++){
						one[0]=num[i];
						one[1]=num[j];
						one[2]=(*it)[0];
						one[3]=(*it)[1];
						sort(one.begin(),one.end());
						int current=one[0];
						int count=1;
						for(int i=1;i<4;i++){
							if(one[i]==current){
								count++;
							}
							else{
								if(count>m[current])
									continue;
								else current=one[i];
								count=1;
							}
						}
						if(count>m[current])continue;
						res.insert(one);
					}
				}
			}
		}
		set<vector<int> >::iterator itt;
		for(itt=res.begin();itt!=res.end();itt++){
			ret.push_back(*itt);
		}
		return ret;
	}
};

转载于:https://www.cnblogs.com/superzrx/p/3330731.html