通过Scanner 类的 next() 与 nextLine() 方法获取输入的字符串,在读取前我们一般需要 使用 hasNext 与 hasNextLine 判断是否还有输入的数据: importjava.util.*;publicclasstest {publicstaticvoidmain(String[] args) { Scanner scan1=newScanner(System.in); System.out.println("next方式接收:");if(scan1.ha...
Scanner scan = new Scanner(System.in); String str = scan.nextLine(); 提示:使用String类和Character类的相应方法来实现。 (2)利用Random类来产生5个20到50之间的随机数。 提示:[n-m] (n,m均为整数,n小于m)之间的随机数的公式为n+(new Random()).nextInt(m-n+1)。 (3)计算从今天算起,100天以...
public static void main(String[] args){ Scanner scan = new Scanner(System.in); System.out.println("输入串(不大于30个字符)。"); String s = scan.nextLine(); Set<Character> sets = new LinkedHashSet<Character>(); removeDuplicate(s,sets); // 去掉重复元素,放入lis check(sets); // 检测...
35.若已定义Scanner scan = new Scanner(System.in);,则以下能从键盘读取一个字符的语句是(A)。 A.char c = (scan.next()).charAt(0); B.char c = scan.nextChar(); C.char c = Character.parseChar(scan.next()); D.char c = (char)(scan.next...
1//通过Map 类实现,通过键值对的方式,可以将输入的字符串的每一个字符,作为键,每个字符出现的次数作为值:如下:23publicclassFind {4publicstaticvoidmain(String[] args){5String scan=newScanner(System.in).nextLine();//获取键盘上输入的字符串;6Map<Character,Integer> map =newHashMap<Character,Integer>(...
An open quote stops at end-of-line unless \ is the last character, which then joins the next line by removing all leading white space characters. Wildcards (*) aren’t allowed in these lists (such as specifying *.java). Use of the at sign (@) to recursively interpret files isn’...
A channel to scan charsetName String The encoding type used to convert bytes from the channel into characters to be scanned Attributes RegisterAttribute Exceptions IllegalArgumentException if the specified character set is not found. Remarks Java documentation for java.util.Scanner.Scanner(java.nio...
next public String next() Finds and returns the next complete token from this scanner. A complete token is preceded and followed by input that matches the delimiter pattern. This method may block while waiting for input to scan, even if a previous invocation of hasNext() returned true. Specif...
publicstaticvoid main<String args[]>{ int x,y; Scanner scan=new Scanner<System.in>; System.out.println<"请输入一个整数:">; x=scan.nextInt<>; y=x%10+x/10%10+x/100%10; System.out.println<x+"各个数位的和="+y>; } } 第三章 知识点汇总 1、 简单 if 结构 if<布尔类型表达式>{...
int userMove = scan.nextInt(); return userMove; } public static void usersMove(String playerName, int gesture) { int userMove = move(); if (userMove == -1) { break; } 通常,更改方法的返回类型与返回语句中的类型一致,可以解决这个问题。例如,下面的void可以改为int: ...