Find the equipment indices
Here is a simple program test task, it doesn't have very diffcult logic:
A zero-indexed array A consisting of N integers is given. An equilibrium index of this array is any integer P such that 0 ≤ P < N and the sum of elements of lower indices is equal to the sum of elements of higher indices, i.e.
A[0] + A[1] + ... + A[P−1] = A[P+1] + ... + A[N−2] + A[N−1].
Sum of zero elements is assumed to be equal to 0. This can happen if P = 0 or if P = N−1.
For example, consider the following array A consisting of N = 7 elements:
A[0] = -7 A[1] = 1 A[2] = 5
A[3] = 2 A[4] = -4 A[5] = 3
A[6] = 0
P = 3 is an equilibrium index of this array, because:
- A[0] + A[1] + A[2] = A[4] + A[5] + A[6]
P = 6 is also an equilibrium index, because:
- A[0] + A[1] + A[2] + A[3] + A[4] + A[5] = 0
and there are no elements with indices greater than 6.
P = 7 is not an equilibrium index, because it does not fulfill the condition 0 ≤ P < N.
Write a function
int solution(int A[], int N);
int solution(NSMutableArray *A);
int solution(const vector<int> &A);
class Solution { int solution(int[] A); }
class Solution { public int solution(int[] A); }
object Solution { def solution(A: Array[Int]): Int }
function solution(A);
function solution(A)
function solution($A);
function solution(A: array of longint; N: longint): longint;
def solution(A)
sub solution { my (@A)=@_; ... }
def solution(a)
Private Function solution ( A As Integer() ) as Integer
that, given a zero-indexed array A consisting of N integers, returns any of its equilibrium indices. The function should return −1 if no equilibrium index exists.
Assume that:
- N is an integer within the range [0..10,000,000];
- each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].
For example, given array A such that
A[0] = -7 A[1] = 1 A[2] = 5
A[3] = 2 A[4] = -4 A[5] = 3
A[6] = 0
the function may return 3 or 6, as explained above.
Complexity:
- expected worst-case time complexity is O(N);
- expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).
Elements of input arrays can be modified.
At very beginning, I have wrotten down this:
// 62 of 100
using System;
// you can also use other imports, for example:
// using System.Collections.Generic;
class Solution {
public int solution(int[] A) {
if(A.Length <=1)
return -1;
//abc
int sumA = 0;
foreach(int i in A){
sumA += i;
} int j = 0;
int tempSum = A[0];
for(j = 1; j<A.Length; j++){ //A[0] + A[1] + ... + A[P−1] = A[P+1] + ... + A[N−2] + A[N−1].
//return P
if(sumA - tempSum - A[j] == tempSum){
break;
}
tempSum += A[j];
}
if (j == A.Length || j == A.Length - 1)
return -1;
else
return j;
}
}
I got 62 points of 100. because it missed border check and one misunderstanding.
So I updated it to this one:
//100 of 100
using System;
// you can also use other imports, for example:
// using System.Collections.Generic;
class Solution {
public int solution(int[] A) {
//N is an integer within the range [0..10,000,000];
if(A.Length <1 || A.Length >10000000)
return -1;
//single number
if(A.Length == 1)
return 0; //each element of array A is an integer within the range [−2,147,483,648..2,147,483,647]
Int64 sumA = 0;
foreach(int i in A){
sumA += i;
} int j = 0; //each element of array A is an integer within the range [−2,147,483,648..2,147,483,647]
Int64 tempSum = 0;
for(; j<A.Length; j++){
//A[0] + A[1] + ... + A[P−1] = A[P+1] + ... + A[N−2] + A[N−1].
//return P
if(sumA - tempSum - A[j] == tempSum){
break;
}
tempSum += A[j];
}
//P = 7 is not an equilibrium index, because it does not fulfill the condition 0 ≤ P < N
if (j == A.Length)
return -1;
else
return j;
}
}
Now, I got 100 points.
This task is from a test website.
Find the equipment indices的更多相关文章
- elasticsearch auto delete old indices
定在crontab 每天执行 crontab -e * 2 * * * ~/autodelete.py Python 代码如下 #!/usr/bin/env python # encoding:utf ...
- Equipment Box[HDU1110]
Equipment Box Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Tot ...
- 转载:监控每个节点(Indices部分)
集群的健康只是一个方面,它是对整个集群所有方面的一个很高的概括.节点状态的api是另外一个方面,它提供了关于你的集群中每个节点令你眼花缭乱的统计数据. 节点的状态提供了那么多的统计数据,在你很熟悉它们 ...
- HDOJ 3177 Crixalis's Equipment
Crixalis's Equipment Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Oth ...
- How do I list all tables/indices contained in an SQLite database
How do I list all tables/indices contained in an SQLite database If you are running the sqlite3 comm ...
- HDU ACM 3177 Crixalis's Equipment
Crixalis's Equipment Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Oth ...
- 杭电 3177 Crixalis's Equipment
http://acm.hdu.edu.cn/showproblem.php? pid=3177 Crixalis's Equipment Time Limit: 2000/1000 MS (Java/ ...
- Hdu 3177 Crixalis's Equipment
Crixalis's Equipment Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Oth ...
- /Home/Tpl/Equipment/rangeIndex.html 里调用魔板
<pre name="code" class="html">demo:/var/www/DEVOPS# vim ./Home/Tpl/Equipme ...
随机推荐
- 关于SCRUM站立会议
查询过后对SCRUM站立会议有了初步的了解 站立会议:在敏捷流程的冲刺阶段中,每一天都会举行项目状况会议,强迫每个人向同伴报告进度,迫使大家把问题摆在明面上,这个会议被称为“scrum”或“每日站立会 ...
- Postgres 9.4 feature highlight: REPLICA IDENTITY and logical replication
Among the many things to say about logical replication features added in PostgreSQL 9.4, REPLICA IDE ...
- [Linux] CentOS 加入开机启动
1.在/etc/init.d/目录下新建一个文件:autostart.sh #!/bin/sh #chkconfig: 2345 80 80 #description: auto start web ...
- 深入浅出WPF开发下载
为什么要学习WPF? 许多朋友也许会问:既然表示层技术那么多,为什么还要推出WPF作为表示层技术呢?我们话精力学习WPF有什么收益和好处呢,这个问题我们从两个方面进行回答. 首先,只要开发表示层程序 ...
- 一、Spring——IoC
IOC概述 Spring中IOC的概念,控制反转概念其实包含两个层面的意思,"控制"是接口实现类的选择控制权:而"反转"是指这种选择控制权从调用者转移到外部第三 ...
- nginx,php相关
nginx安装 http://www.nginx.cn/install php安装 https://segmentfault.com/a/1190000004123048#articleHeader5 ...
- 命令查看java的class字节码文件、verbose、synchronize、javac、javap
查看Java字节码 1 javac –verbose查看运行类是加载了那些jar文件 HelloWorld演示: public class Test { public static void main ...
- ios开发 iphone中获取网卡地址和ip地址
这是获取网卡的硬件地址的代码,如果无法编译通过,记得把下面的这几个头文件加上把. #include <sys/socket.h> // Per msqr#include <sys/s ...
- CyclicBarrier类合唱演绎
package a.jery; import java.util.concurrent.CyclicBarrier; import java.util.concurrent.ExecutorServi ...
- oracle字符集相关问题
整理自网络+实验 字符集介绍 影响Oracle数据库字符集最重要的参数是NLS_LANG参数. 它的格式如下: NLS_LANG = language_territory.charset NLS_L ...