1、编写一个控制台应用程序,接收一个长度大于3的字符串,完成 下列功能:

2025-03-27 23:49:50
推荐回答(2个)
回答1:

01.using System;
02.class Test6
03.{
04. public static void Main()
05. {
06. string str = "";
07. while (str.Length <= 3)
08. {
09. Console.Write("请输入一个长度大于3的字符串:");
10. str = Console.ReadLine();
11. }
12. //(1)
13. Console.WriteLine("字符串的长度为:{0}", str.Length);
14. //(2)
15. int i = str.IndexOf('a');
16. if (i > -1)
17. {
18. Console.WriteLine("第一个出现字母a的位置是:{0}", i);
19. }
20. else
21. {
22. Console.WriteLine("字符串中不包含字母a。");
23. }
24. //(3)
25. string str1 = str.Insert(3, "hello"); //在第3个(初始序号为)字符前插入hello
26. Console.WriteLine("插入hello后的结果为:{0}", str1);
27. //(4)
28. string str2 = str1.Replace("hello", "me");
29. Console.WriteLine("将hello替换为me后的结果为:{0}", str2);
30. //(5)
31. string[] arr = str2.Split('m');
32. Console.WriteLine("以m为分隔符分离后的字符串有:");
33. for (int j = 0; j < arr.Length; j++)
34. {
35. Console.WriteLine(arr[j]);
36. }
37. }
38.}

回答2:

public class ST {
public static void main(String[] args) {
String str = "hello world!";

int len = str.length();
System.out.println("字符串长度为:"+len);

StringBuffer sb = new StringBuffer(str);
StringBuffer strnew = sb.insert(3, "hello");
String s = strnew.toString();
System.out.println("字符串第一次改变后:"+s);

String s1 = s.replace("hello", "me");
System.out.println("字符串替换后:"+s1);

int lenth = s1.indexOf("m");
String s2 = s1.substring(0,lenth);
String s3 = s1.substring(lenth);
System.out.println("m前得字符为:"+s2);
System.out.println("m后的字符为:"+s3);
}

}
输出结果为:
字符串长度为:12
字符串第一次改变后:helhellolo world!
字符串替换后:helmelo world!
m前得字符为:hel
m后的字符为:melo world!

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