LeetCode 3Sum Closest 最接近目标数的三个数和

最新推荐文章于 2025-07-20 10:41:25 发布

最新推荐文章于 2025-07-20 10:41:25 发布 · 191 阅读

本文探讨了如何在给定整数数组中找到三个数的和，使得该和与目标值之间的差距最小。通过算法实现，解决了与经典三数之和问题相似但又略有不同的挑战。

3Sum Closest

Given an arraySofnintegers, find three integers inSsuch that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.

    For example, given array S = {-1 2 1 -4}, and target = 1.

    The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

这道题和3Sum差不多，不过也有不一样的，主要是：

1 这里不用判断处理重复问题

2 要比较其中的三个数的和与目标数的差的大小。

class Solution {
public:
	int threeSumClosest(vector<int> &num, int target)
	{
		switch (num.size())
		{
		case 0: 
			return 0;
		case 1:
			return num[0];
		case 2:
			return num[0] + num[1];
		default:
			break;
		}
		sort(num.begin(), num.end());
		int closet = 0;
		int sum = 0;
		int i = 0, j = 0, k = num.size()-1;
		int diff = INT_MAX;

		for (i = 0; i < k-1; i++)
		{
			for (j = i+1; j < k;)
			{
				sum = num[i] + num[j] + num[k];
				if (sum == target)
				{
					return sum;
				}
				if (abs(sum-target) < diff)
				{
					closet = sum;
					diff = abs(sum - target);
				}
				if (sum < target)
				{
					j++;
				}
				else
				{
					k--;
				}
			}
		}
		return closet;
	}
};