Java 字符串子字符串错误
问题:Java 字符串子字符串错误,java,string,Java,String,问题: Input:-gandhi output:- Gandhi Input:-mahatma gandhi output:- M. Gandhi Input:-Mohndas Karamchand ganDhi output:- M. K. Gandhi 答复: public class Chef_NITIKA { static Scanner scan=new Scanner(System.in);
Input:-gandhi output:- Gandhi
Input:-mahatma gandhi output:- M. Gandhi
Input:-Mohndas Karamchand ganDhi output:- M. K. Gandhi
答复:
public class Chef_NITIKA {
static Scanner scan=new Scanner(System.in);
public static void main(String[] args) {
String name= myString();
System.out.println("nam is :"+name);
mySformatter(name);
}
private static void mySformatter(String name) {
int count=0;
for(char c:name.toCharArray()){
if(c==' '){
count+=1;
}
}
System.out.println(count+" blank spaces");
if(count==0){
char ch=name.charAt(0);
name=Character.toUpperCase(ch)+name.substring(1);
System.out.println("nam is :"+name);
}
else if(count==1){
char ch=name.charAt(0);
name= name.replace(' ', '.');
String subname=name.substring(name.indexOf(".")+1);
char c=subname.charAt(0);
subname=Character.toUpperCase(c)+subname.substring(1);
name=Character.toUpperCase(ch)+"."+subname;
System.out.println("nam is :"+name);
}
else if(count==2){
char ch=name.charAt(0);
// name= name.replace(' ', '.');
String subname=name.substring(name.indexOf(" ")+1);
System.out.println(subname);
String subsubname=subname.substring(name.indexOf(" "));
System.out.println(subsubname);
char c=subname.charAt(0);
char c1=subsubname.charAt(0);
subname=Character.toUpperCase(c)+subname.substring(1);
name = Character.toUpperCase(ch)+"."+Character.toUpperCase(c)+"."+Character.toUpperCase(c1)+subsubname.substring(1);
System.out.println("nam is :"+name);
}
}
private static String myString() {
System.out.println("enter the string");
String s=scan.nextLine();
StringBuffer name=new StringBuffer();
// name.append(s);
return s;
}
}
当我输入“abc cde fgh”时,我没有得到所需的输出,我得到的输出是A.C..fgh
有什么有效的方法来解决这个问题吗
不期望的输出:-
enter the string
iam writing onStack
nam is :iam writing onStack
2 blank spaces
writing onStack
ing onStack
nam is :I.W.Ing onStack
我希望输出为I.W.OnStack只需将名称拆分为组件,然后形成所需的缩写:
public static String getName(String input) {
String[] names = input.split("\\s+");
String output = "";
// replace 1st through second to last names with capitalized 1st letter
for (int i=names.length; i > 1; --i) {
output += names[names.length - i].substring(0, 1).toUpperCase() + ". ";
}
// append full last name, first letter capitalized, rest lower case
output += names[names.length - 1].substring(0, 1).toUpperCase()
+ names[names.length - 1].substring(1).toLowerCase();
return output;
}
public static void main(String[] args) {
System.out.println(getName("gandhi"));
System.out.println(getName("mahatma gandhi"));
System.out.println(getName("Mohndas Karamchand ganDhi"));
}
输出:
Gandhi
M. Gandhi
M. K. Gandhi
此处演示:
Gandhi
M. Gandhi
M. K. Gandhi
更新:
Gandhi
M. Gandhi
M. K. Gandhi
下面是一个例子,我已经纠正了您的原始代码,至少部分是这样。问题如下:
String subsubname = subname.substring(name.indexOf(" "));
我将其改为:
String subsubname = subname.substring(subname.indexOf(" ") + 1);
您没有正确识别名称第三部分的第一个字符。话虽如此,您当前的方法冗长、难以理解且缺乏灵活性。实际上,您可能希望使用更精简的方法来解决此问题。只需将名称拆分为组件,然后形成所需的缩写:
public static String getName(String input) {
String[] names = input.split("\\s+");
String output = "";
// replace 1st through second to last names with capitalized 1st letter
for (int i=names.length; i > 1; --i) {
output += names[names.length - i].substring(0, 1).toUpperCase() + ". ";
}
// append full last name, first letter capitalized, rest lower case
output += names[names.length - 1].substring(0, 1).toUpperCase()
+ names[names.length - 1].substring(1).toLowerCase();
return output;
}
public static void main(String[] args) {
System.out.println(getName("gandhi"));
System.out.println(getName("mahatma gandhi"));
System.out.println(getName("Mohndas Karamchand ganDhi"));
}
/*
* To change this license header, choose License Headers in Project Properties.
* To change this template file, choose Tools | Templates
* and open the template in the editor.
*/
package stack;
import java.util.Scanner;
import java.util.StringTokenizer;
/**
*
* @author xxz
*/
public class NAmes {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
final String name = sc.nextLine();
System.out.println(formatedString(name));
sc.close();
}
private static String formatedString(String name) {
if(name!=null && name.length()!=0){
final StringTokenizer tokenizer = new StringTokenizer(name, " ");
StringBuilder result = new StringBuilder("");
while(tokenizer.hasMoreTokens()){
StringBuilder currentToken =new StringBuilder(tokenizer.nextToken());
if(!tokenizer.hasMoreTokens()){
result.append(currentToken.toString().toUpperCase());
} else{
result.append(currentToken.toString().toUpperCase().charAt(0)+". ");
}
}
return result.toString();
}
return "";
}
}
输出:
Gandhi
M. Gandhi
M. K. Gandhi
此处演示:
Gandhi
M. Gandhi
M. K. Gandhi
更新:
Gandhi
M. Gandhi
M. K. Gandhi
下面是一个例子,我已经纠正了您的原始代码,至少部分是这样。问题如下:
String subsubname = subname.substring(name.indexOf(" "));
我将其改为:
String subsubname = subname.substring(subname.indexOf(" ") + 1);
您没有正确识别名称第三部分的第一个字符。话虽如此,您当前的方法冗长、难以理解且缺乏灵活性。实际上,您可能希望使用更精简的方法来解决此问题。@Rushabh Oswal您可以通过以下方法获得所需的结果
/*
* To change this license header, choose License Headers in Project Properties.
* To change this template file, choose Tools | Templates
* and open the template in the editor.
*/
package stack;
import java.util.Scanner;
import java.util.StringTokenizer;
/**
*
* @author xxz
*/
public class NAmes {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
final String name = sc.nextLine();
System.out.println(formatedString(name));
sc.close();
}
private static String formatedString(String name) {
if(name!=null && name.length()!=0){
final StringTokenizer tokenizer = new StringTokenizer(name, " ");
StringBuilder result = new StringBuilder("");
while(tokenizer.hasMoreTokens()){
StringBuilder currentToken =new StringBuilder(tokenizer.nextToken());
if(!tokenizer.hasMoreTokens()){
result.append(currentToken.toString().toUpperCase());
} else{
result.append(currentToken.toString().toUpperCase().charAt(0)+". ");
}
}
return result.toString();
}
return "";
}
}
私有静态void mySformatter(字符串名称){
字符串[]tempArray=name.split(“”);
StringBuilder outputStr=新建StringBuilder();
int len=tempArray.length;
对于(inti=0;i@Rushabh Oswal),您可以通过以下方法获得所需的结果
私有静态void mySformatter(字符串名称){
字符串[]tempArray=name.split(“”);
StringBuilder outputStr=新建StringBuilder();
int len=tempArray.length;
对于(int i=0;i“abc cde fgh”的期望输出是什么?“abc cde fgh”的期望输出是什么?我编译了你的代码并得到了正确的输出,但是你能在我的代码中找到错误吗?主要是当计数为2ya时,我已经发布了我的输出getting@RushabhOswal我现在正在测试您的代码,请给我几分钟时间。在ny second演示中,我能够获得abc cde fgh
的正确输出。正如我所说,您当前的方法是不理想,所以我不会花更多时间调试它。我编译了你的代码并得到了正确的输出,但是你能在我的代码中找到错误吗?主要是当计数为2ya时,我已经发布了我的输出getting@RushabhOswal我现在正在测试您的代码,请给我几分钟时间。在第二次演示中,我能够获得abc cde的正确输出fgh
。正如我所说,您当前的方法并不理想,因此我不会花更多时间调试它。我想知道当计数器==2时,代码中的错误是什么。我想知道当计数器==2时,代码中的错误是什么