LeetCode 1533. Find the Index of the Large Integer(二分查找)「建议收藏」

LeetCode 1533. Find the Index of the Large Integer(二分查找)「建议收藏」文章目录1.题目2.解题1.题目Wehaveanintegerarrayarr,wherealltheintegersinarrareequalexceptforoneintegerwhichislargerthantherestoftheintegers.Youwillnotbegivendirectaccesstothearray,instead,youwillhaveanAPIArrayReaderwhi

大家好,又见面了,我是你们的朋友全栈君。如果您正在找激活码,请点击查看最新教程,关注关注公众号 “全栈程序员社区” 获取激活教程,可能之前旧版本教程已经失效.最新Idea2022.1教程亲测有效,一键激活。

Jetbrains全系列IDE使用 1年只要46元 售后保障 童叟无欺

文章目录

1. 题目

We have an integer array arr, where all the integers in arr are equal except for one integer which is larger than the rest of the integers. You will not be given direct access to the array, instead, you will have an API ArrayReader which have the following functions:

  • int compareSub(int l, int r, int x, int y): where 0 <= l, r, x, y < ArrayReader.length(), l <= r and x <= y. The function compares the sum of sub-array arr[l…r] with the sum of the sub-array arr[x…y] and returns:
    1 if arr[l]+arr[l+1]+...+arr[r] > arr[x]+arr[x+1]+...+arr[y].
    0 if arr[l]+arr[l+1]+...+arr[r] == arr[x]+arr[x+1]+...+arr[y].
    -1 if arr[l]+arr[l+1]+...+arr[r] < arr[x]+arr[x+1]+...+arr[y].
  • int length(): Returns the size of the array.

You are allowed to call compareSub() 20 times at most. You can assume both functions work in O(1) time.

Return the index of the array arr which has the largest integer.

Follow-up:

  • What if there are two numbers in arr that are bigger than all other numbers?
  • What if there is one number that is bigger than other numbers and one number that is smaller than other numbers?
Example 1:
Input: arr = [7,7,7,7,10,7,7,7]
Output: 4
Explanation: The following calls to the API
reader.compareSub(0, 0, 1, 1) 
// returns 0 this is a query comparing the sub-array (0, 0) with the sub array (1, 1), (i.e. compares arr[0] with arr[1]).
Thus we know that arr[0] and arr[1] doesn't contain the largest element.
reader.compareSub(2, 2, 3, 3) 
// returns 0, we can exclude arr[2] and arr[3].
reader.compareSub(4, 4, 5, 5) 
// returns 1, thus for sure arr[4] is the largest element in the array.
Notice that we made only 3 calls, so the answer is valid.

Example 2:
Input: nums = [6,6,12]
Output: 2
 
Constraints:
2 <= arr.length <= 5 * 10^5
1 <= arr[i] <= 100
All elements of arr are equal except for one element which is larger than all other elements.

来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/find-the-index-of-the-large-integer
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。

2. 解题

/** * // This is the ArrayReader's API interface. * // You should not implement it, or speculate about its implementation * class ArrayReader { * public: * // Compares the sum of arr[l..r] with the sum of arr[x..y] * // return 1 if sum(arr[l..r]) > sum(arr[x..y]) * // return 0 if sum(arr[l..r]) == sum(arr[x..y]) * // return -1 if sum(arr[l..r]) < sum(arr[x..y]) * int compareSub(int l, int r, int x, int y); * * // Returns the length of the array * int length(); * }; */

class Solution { 
   
public:
    int getIndex(ArrayReader &reader) { 
   
        int n = reader.length(), l = 0, r = n-1, midl, midr;
        int flag;
        while(l < r)
        { 
   
        	if((r-l)&1)//差为奇数
        		midl = (l+r)/2,
        		midr = (l+r)/2+1;
        	else//偶数
        		midl = midr = (l+r)/2;
        	flag = reader.compareSub(l, midl, midr,r);
        	if(flag > 0)//左边和大
        		r = midl;
        	else if(flag < 0)//右边和大
        		l = midr;
        	else//相等找到了
        		return midl;
        }
        return l;
    }
};

200 ms 39.7 MB


我的CSDN博客地址 https://michael.blog.csdn.net/

长按或扫码关注我的公众号(Michael阿明),一起加油、一起学习进步!
Michael阿明

版权声明:本文内容由互联网用户自发贡献,该文观点仅代表作者本人。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容, 请发送邮件至 举报,一经查实,本站将立刻删除。

发布者:全栈程序员-用户IM,转载请注明出处:https://javaforall.cn/160104.html原文链接:https://javaforall.cn

【正版授权,激活自己账号】: Jetbrains全家桶Ide使用,1年售后保障,每天仅需1毛

【官方授权 正版激活】: 官方授权 正版激活 支持Jetbrains家族下所有IDE 使用个人JB账号...

(0)


相关推荐

  • Linux——常用命令(压缩和解压缩)

    Linux——常用命令(压缩和解压缩)在Linux中可以识别的常见压缩格式有几十种,比如.zip、.gz、.bz2、.tar、.tar.gz、.tar.bz2等。1、zip格式

  • svn服务端安装使用教程_ug安装教程

    svn服务端安装使用教程_ug安装教程直接上干货第一步:首先要下载SVN,记住是服务端哦,客户端的安装请查看:https://blog.csdn.net/u012974916/article/details/116002250下载地址:https://www.visualsvn.com/server/download/,根据系统选择对应的版本。第二步:双击安装程序VisualSVN-Server-4.2.1-x64.msi第三步:勾选复选框选择同意,然后选择Next,选择Upgrade第四步:设置服..

    2022年10月18日
  • TLSF算法分析

    TLSF算法分析注:本文的大部分内容摘录自论文《TLSF:aNewDynamicMemoryAllocatorforReal-TimeSystems》,可以通过“科学上网”访问如下链接阅读原文:http://www.gii.upv.es/tlsf/files/ecrts04_tlsf.pdf。什么是TLSFTLSF是TwoLevelSegregatedFitmemoryal

  • Mysql和redis_简述Redis和MySQL的区别[通俗易懂]

    Mysql和redis_简述Redis和MySQL的区别[通俗易懂]我们知道,mysql是持久化存储,存放在磁盘里面,检索的话,会涉及到一定的IO,为了解决这个瓶颈,于是出现了缓存,比如现在用的最多的memcached(简称mc)。首先,用户访问mc,如果未命中,就去访问mysql,之后像内存和硬盘一样,把数据复制到mc一部分。redis和mc都是缓存,并且都是驻留在内存中运行的,这大大提升了高数据量web访问的访问速度。然而mc只是提供了简单的数据结构,比如…

  • unity开发微信小游戏1[通俗易懂]

    unity开发微信小游戏1[通俗易懂]unity开发微信小游戏

发表回复

您的电子邮箱地址不会被公开。

关注全栈程序员社区公众号