LeetCode 4Sum-优快云博客

本文详细介绍了如何解决四数之和问题，即在一个整数数组中找到四个元素，使得它们的和等于给定的目标值。文章通过提供一个C++程序实例，展示了如何使用排序和双指针技巧来高效地找到所有符合条件的四元组，并确保不包含重复的组合。

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Given an arraySofnintegers, are there elementsa,b,c, anddinSsuch thata+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)

这道题和3Sum的思路是一样的。时间复杂度是O(n^3)，增加了一个外循环判断，程序长了点，总的程序结构也是一样的。

#include<iostream>
#include<vector>
#include<string>
#include<algorithm>

using namespace std;

class Solution {
public:
	vector<vector<int> > fourSum(vector<int> &num, int target) 
	{
		vector<vector<int> > result;
		if (num.size() < 4)
		{
			return result;
		}
		vector<int> fNum(4);
		sort(num.begin(), num.end());

		int i = 0, j = 0, k = 0, r = num.size()-1;
		while (i < r-2)
		{
			for (j = i+1; j < r-1;)
			{
				//Notice: don't forget to reset the value of r = num.size()-1;
				for (k = j+1, r = num.size()-1; k < r; )
				{
					int temp = num[i] + num[j] + num[k] + num[r];
					if (temp == target)
					{
						fNum[0] = num[i]; fNum[1] = num[j]; fNum[2] = num[k]; fNum[3] = num[r];
						result.push_back(fNum);
						k++; r--;
						//注意判断重复的时候需要放进判断条件括号里面，因为没有移动的时候不需要判断
						//k和r都不一定每次移动的
						while (k<r && num[k] == num[k-1])
						{
							k++;
						}
						while (k<r && num[r] == num[r+1])
						{
							r--;
						}
					}
					else if (temp < target)
					{
						k++;
						while (k<r && num[k] == num[k-1])
						{
							k++;
						}
					}
					else
					{
						r--;
						while (k<r && num[r] == num[r+1])
						{
							r--;
						}
					}

				}
				j++;
				while (j<r-1 && num[j] == num[j-1])
				{
					j++;
				}
			}
			i++;
			while (i<r-2 && num[i] == num[i-1])
			{
				i++;
			}
		}
		return result;
	}
};
int main() 
{
	int a[] = {2,7,9,5,5,3,8,10,-1,-2,-5,-6,-17,-35};
	int len = sizeof(a)/sizeof(int);
	vector<int> vi(a, a+len);
	cout<<"The array is:\n";
	for (auto x:vi)
		cout<<x<<" ";
	cout<<endl;

	Solution solu;
	vector<vector<int> > vt = solu.fourSum(vi, 10);

	for (auto x:vi)
		cout<<x<<" ";
	cout<<endl;
	for (auto x:vt)
	{
		for (auto y:x)
			cout<<y<<" ";
		cout<<endl;
	}

	system("pause");
	return 0;
}

12.13update：更新使用set避免重复。程序更加简洁。

class Solution {
public:
	vector<vector<int> > fourSum(vector<int> &num, int target)
	{
		int n = num.size()-1;
		if (n <3) return vector<vector<int> >();
		set<vector<int> > svi;
		sort(num.begin(), num.end());
		vector<int> vi(4);
		for (int i = 0; i < n-2; i++)
		{
			for (int j = i+1; j < n-1; j++)
			{
				for (int k = j+1, n = num.size()-1; k < n;)
				{
					int t = num[i]+num[j]+num[k]+num[n];
					if (t == target)
					{
						vi[0] = num[i]; vi[1] = num[j]; vi[2] = num[k]; vi[3] = num[n];
						svi.insert(vi);
						k++; n--;
					}
					else if (t < target)
						k++;
					else n--;
				}//k
			}//j
		}//for i
		return vector<vector<int> >(svi.begin(), svi.end());
	}
};

//2014-1-25
class Solution {
public:
	vector<vector<int> > fourSum(vector<int> &num, int target) 
	{
		sort(num.begin(), num.end());
		vector<int> tmp(4);
		vector<vector<int> > rs;
		//i<d-2; j<d-1可以让程序由原来的444,432ms左右优化到240,232ms
		int i = 0, j = 1, k = 2, d = num.size()-1;

		for (; i < d-2; i++)
		{
			for (j = i+1; j < d-1; j++)
			{
				for (k = j+1, d = num.size()-1; k < d; )
				{
					int sum = num[i]+num[j]+num[k]+num[d];
					if (sum > target) d--;
					else if (sum < target) k++;
					else
					{
						tmp[0] = num[i];
						tmp[1] = num[j];
						tmp[2] = num[k];
						tmp[3] = num[d];
						rs.push_back(tmp);
						k++;
						d--;
						while (k<d && num[k]==num[k-1]) k++;
						while (k<d && num[d]==num[d+1]) d--;
					}
					while (j<d-1 && num[j]==num[j+1]) j++;
				}
				while (i<d-2 && num[i]==num[i+1]) i++;
			}
		}
		return rs;
	}
};