CareerCup Given an array having positive integers, find a subarray which adds to a given number

本文介绍了一种在正整数数组中找到连续子数组使其和等于给定数值的方法。该算法采用一次遍历的方式,通过调整运行总和来确保效率达到O(n)级别。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

Given an array having positive integers, find a continous subarray which adds to a given number.

-----------------------------------------------


Take notice that all numbers are positive!

It can be done O(n). 
My logic is as follows:- Keep adding elements to a running sum till it is either equal to given number or is greater than the given number. As soon as we add a number and the running number is greater than the given number we subtract the elements from the beginning of the array till it is again less than the given number. After the current sum is less than the given number we add the next element and check and continue the same process.

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