Leet Code Medium 16 3Sum Closest

本文介绍了一种寻找数组中三个整数之和最接近给定目标值的有效算法。利用双指针技术,时间复杂度达到O(n²)。通过实例演示了如何找到最接近目标的三数之和。

Given an array S of n integers, find three integers in S such 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).

Analysis

This problem is similar to 2 Sum. This kind of problem can be solved by using a similar approach, i.e., two pointers from both left and right.

Java Solution

public int threeSumClosest(int[] nums, int target) {
    int min = Integer.MAX_VALUE;
	int result = 0;
 
	Arrays.sort(nums);
 
	for (int i = 0; i < nums.length; i++) {
		int j = i + 1;
		int k = nums.length - 1;
		while (j < k) {
			int sum = nums[i] + nums[j] + nums[k];
			int diff = Math.abs(sum - target);
 
			if(diff == 0) return sum;
 
			if (diff < min) {
				min = diff;
				result = sum;
			}
			if (sum <= target) {
				j++;
			} else {
				k--;
			}
		}
	}
 
	return result;
}


Time Complexity is O(n^2).

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值