LeetCode 4Sum

最新推荐文章于 2025-08-20 09:37:41 发布

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本文探讨了如何解决四数之和的问题，即在给定的整数数组中找到四个元素，使它们的和等于目标值。文章提供了一个详细的算法实现，并通过C++代码展示了如何避免重复解和确保解决方案集的有效性。

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题目

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)

循环求之前的3SUM问题，n^3

代码：

class Solution {
public:
    vector<vector<int> > fourSum(vector<int> &num, int target) {
		vector<vector<int>> ans;
		int size=num.size();
		if(size<4)
			return ans;
		sort(num.begin(),num.end());
		long long sum;
		vector<int> qua;
		int i=0,j=1,k=2,l=size-1;
		while(k<l)
		{
			while(k<l)	//3SUM
			{
				while(k<l)	//2SUM
				{
					sum=num[i]+num[j]+num[k]+num[l];
					if(sum==target)
					{
						qua.clear();
						qua.push_back(num[i]);
						qua.push_back(num[j]);
						qua.push_back(num[k]);
						qua.push_back(num[l]);
						ans.push_back(qua);
						k++;
						while(k<l&&num[k]==num[k-1])
							k++;
						l--;
						while(k<l&&num[l]==num[l+1])
							l--;
					}
					else if(sum<target)
						k++;
					else
						l--;
				}
				j++;
				while(j<size&&num[j]==num[j-1])
					j++;
				k=j+1;
				l=size-1;
			}
			i++;
			while(i<size&&num[i]==num[i-1])
				i++;
			j=i+1;
			k=i+2;
			l=size-1;
		}
		return ans;
    }
};