LeetCode:3Sum Closest
2016-06-15 17:18
423 查看
3Sum Closest
Total Accepted: 80994 TotalSubmissions: 275223 Difficulty: Medium
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).
Subscribe to see which companies asked this question
Hide Tags
Array Two
Pointers
Hide Similar Problems
(M) 3Sum (M)
3Sum Smaller
思路:
暴力解法是三个循环i,j,k,判断(nums[i] +
nums[i] + nums[k])与target的最接近值,时间复杂度O(n^3);
但是其实有许多重复的计算,就是在三个指针交叉指向的时候;
如:(i = 0,j=2,k=4)与(i=2,j=4,k=2)这些都是不必要的。
三个指针,first loop(0~n-1),每次seond(从first+1开始)和third(从n-1开始)相互逼近,这样可以去掉重复值;
时间复杂度可降到O(n^2)。
java code:
public class Solution { public int threeSumClosest(int[] nums, int target) { int n = nums.length; int closetSum = nums[0] + nums[1] + nums[n - 1]; Arrays.sort(nums); for(int first = 0; first < n-2; first++) { //不同有两个指针指向同一位置,因此是n-2 //if(first > 0 && nums[first] == nums[first-1]) continue;//去重用,这里不需要 int second = first + 1, third = n - 1; while(second < third) { int curSum = nums[first] + nums[second] + nums[third]; if(curSum==target) return target; if(Math.abs(target - curSum) < Math.abs(target - closetSum)) closetSum = curSum; if(curSum < target) second++; else third--; } } return closetSum; } }
相关文章推荐
- 关于go1.6使用vendor的坑
- [Java] 使用多个 if 语句、else if 和 switch 的区别
- CSS3鼠标滑过彩色按钮动画特效
- 什么是死锁?如何避免死锁?
- yum安装PHP/yum升级PHP
- Linux vim编辑器简单使用之一:vim编辑器的几种模式
- 未能加载文件或程序集“log4net, Version=1.2.10.0, Culture=neutral, PublicKeyToken=1b44e1d426115821”或它的某一个依赖项。系统找不到指定的文件。
- 影响postgresql性能的几个重要参数
- windows下go编码转换问题
- [Java] 类和方法及调用
- Apache中Httpd.conf详解
- ionic 卡片滑动效果
- linux make: *** No targets specified and no makefile found. Stop.
- IE9 不F12打开控制台,代码不执行。打开后正常
- iOS RunLoop
- 深入浅析JS的数组遍历方法(推荐)
- Permutations II
- 集中采购业务的实现
- ios学习路线—Objective-C(Category)
- 解压本地zip,读取数据更新UI