求数组三项之和最接近某个目标数字

最新推荐文章于 2022-02-24 21:10:14 发布

转载最新推荐文章于 2022-02-24 21:10:14 发布 · 197 阅读

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原文链接：http://www.cnblogs.com/lilei2blog/p/7838554.html

文章标签：

#java

本文介绍了一种寻找数组中三个整数使得其和最接近给定目标值的有效算法。通过先对数组进行排序，然后使用双指针技术遍历数组来逼近目标和，实现了O(n²)的时间复杂度。

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描述

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).

思路

1、排序数组

2、循环遍历数组，两面夹逼

空间复杂度O(1)，时间复杂度O(n2)

代码

package com.lilei.spring_boot_db.boot.pack1115;

import java.util.Arrays;

public class sum3closest {

	public static void main(String[] args) {

		int[] array = new int[]{1,2,3,4,5,6,7,9};
		
		System.out.println(sum(array,25));
	}

	static int sum(int[] array, int target) {
		int result = Integer.MIN_VALUE;

		Arrays.sort(array);

		for (int i = 0; i < array.length; i++) {

			int left = 0;
			int right = array.length - 1;

			while (left < right) {
				if (left != i && right != i) {
					if (Math.abs(target - result) > Math.abs(target
							- (array[left] + array[i] + array[right])))
						result = array[left] + array[i] + array[right];

					if (array[left] + array[i] + array[right] < target) {
						left++;
					} else if (array[left] + array[i] + array[right] > target) {
						right--;
					} else {
						return array[left] + array[i] + array[right];
					}
				} else if (left == i)
					left++;
				else
					right--;
			}

		}

		return result;
	}

}

转载于:https://www.cnblogs.com/lilei2blog/p/7838554.html