C. Magic Odd Square time limit per test:1 second memory limit per test:256 megabytes input:standard input output:standard output Find an n × n matrix with different numbers from 1 to n2, so the sum in each row, column and both main diagonals are odd.
要求很简单,输入一个链表,反转链表后,输出新链表的表头. 反转链表是有2种方法(递归法,遍历法)实现的,面试官最爱考察的算法无非是斐波那契数列和单链表反转,递归方法实现链表反转比较优雅,但是对于不了解递归的同学来说还是有理解难度的. 递归法 总体来说,递归法是从最后一个Node开始,在弹栈的过程中将指针顺序置换的. 为了方便理解,我们以 1->2->3->4这个链表来做演示.输出的效果是4->3->2->1 首先定义Node: public static class
Find an n × n matrix with different numbers from 1 to n2, so the sum in each row, column and both main diagonals are odd. Input The only line contains odd integer n (1 ≤ n ≤ 49). Output Print n lines with n integers. All the integers should be differ
Singleton指的是仅仅被实例化一次的类,比如唯一的系统组件等,成为Singleton的类测试起来也比较困难. 常用的方法: 1.公有静态final域+私有构造器 public class Egg{ public static final Egg INSTANCE=new Egg(); private Egg(){} } 2.私有静态final域,私有构造器+公有静态方法 public class Egg{ private static final Egg INSTANCE=new Egg(