用java编程统计用户从键盘输入的字符串中所包含的字母,数字和其他字符串的个数?

Answer 1

import java.util.Scanner;
public class Test {
 public static void main(String [] args){
  Scanner input = new Scanner(System.in);
  System.out.print("输入字符串：");
  String strs = input.next();
  
  int number = 0;
  int chara = 0;
  int other = 0;
  char [] chs = strs.toCharArray();
  for(char c : chs){
   if(c >= '0' && c <= '9'){
    number++;
   }else if(c >= 'a' && c <= 'z' || c >= 'A' && c <= 'Z' ){
    chara++;
   }else{
    other++;
   }
  }
  System.out.println("数字有:" + number + "个，字符有" + chara + "个，其他有：" + other + "个。" );
 }
}

Answer 2

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class _1 {

/**
* @param args
* @throws IOException
*/
public static void main(String[] args) throws IOException {
// TODO Auto-generated method stub

String zifuchuan = new String("");
int hanzishu = 0;
int zimu = 0;
int kongge = 0;
int shuzi = 0;
int qita = 0;
System.out.print("请输入一行字符：");
BufferedReader stdin = new BufferedReader(new InputStreamReader(
System.in));
zifuchuan = stdin.readLine();
byte[] bytes = zifuchuan.getBytes();
for (int i = 0; i < bytes.length; i++) {
if ((bytes[i] >= 65 && bytes[i] <= 90)
|| (bytes[i] >= 97 && bytes[i] <= 122))
zimu++;
else if (bytes[i] == 32)
kongge++;
else if (bytes[i] >= 48 && bytes[i] <= 57)
shuzi++;
else if (bytes[i] < 0)
hanzishu++;
else
qita++;
}
System.out.println("字符串所占字节个数为：" + bytes.length);
System.out.println("汉字个数为：" + hanzishu / 2);
System.out.println("英文字母个数为：" + zimu);
System.out.println("空格个数为：" + kongge);
System.out.println("数字个数为：" + shuzi);
System.out.println("其他字符个数为：" + qita);
}

}

Answer 3

很简单的 用for循环遍历判断