JAVA语言程序设计课后习题----第六单元解析（仅供参考）

1　　本题就是基本函数的用法

import java.util.Scanner;

public class Poone {


    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        System.out.println("请输入一个字符串:");
        String string = input.next();
        System.out.println("显示这个字符串:");
        System.out.println(string);
        System.out.println("这个字符串长度为:");
        System.out.println(string.length());
        System.out.println("该字符串第一个字符");
        System.out.println(string.charAt(0));
        System.out.println("该字符串最后一个字符:");
       // System.out.println(string.charAt(2));
        System.out.println(string.charAt(string.length()-1));
    }

}

2　　String类中有一个方法 public boolean contains（Sting s）就是用来判断当前字符串是否含有参数指定的字符串

import java.util.Scanner;

public class Potwo {
    public static void main(String[] args) {

        Scanner input = new Scanner(System.in);
        System.out.println("请你输入第一个字符串:");
        String s1 = input.next();
        System.out.println("请你输入第二个字符串");
        String s2 = input.next();
        System.out.print("第二个字符串是否为第一个字符串的子串:");
        System.out.println(s1.contains(s2));

    }

}

3　　判断书写格式是否合法，自行百度正则表达式

import java.util.Scanner;

public class Pothree {
    public static void main(String[] args) {
        String regex = "\d{3}+-\d+-\d{3}+-\d{5}+-\d";
        //String a = "887-7-111-50690-4";
        System.out.println("请你输入一个字符串:");
        Scanner input = new Scanner(System.in);
        String a =input.next();
        if (a.matches(regex))
            System.out.println("该字符串合法");
        else
            System.out.println("该字符串不合法");
    }
}

4　　定义一个在字母范围的条件即可

import java.util.Scanner;

public class Pofour {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        System.out.println("请你输入一串字符串:");
        String s = input.next();
        System.out.print("字符串中字母的个数为:"+ countLetters(s));
    }
    public static int countLetters(String s){
        int count =0;
//        字符串转数组
        char []A = s.toCharArray();
        for (int i = 0; i <A.length ; i++) {
            if (A[i]>='a'&&A[i]<='z'||A[i]>='A'&&A[i]<='Z')
                count++;
        }
        return count;
    }
}

5　　本题只要根据题目意思理解就行

public class jinzhi {
    public static String toBinary(int value){
       String s ="";
        while (value!=0){
           s=value%2+s;
            value /= 2;
        }

        return s;
    }

    public static void main(String[] args) {
        System.out.println(toBinary(42));
    }

}

6　　把字符串变成数组，然后根据冒泡法进行排序即可

public class Posix {
    public static void main(String[] args) {

        System.out.println(sort("morning"));
    }
    public static String sort(String s){
        String string=new String();

        char []a=s.toCharArray();
        for (int i = 0; i <a.length-1 ; i++) {
            for (int j = i+1; j < a.length; j++) {
                if (a[i]>a[j]){
                    a[i]^=a[j];
                    a[j]^=a[i];
                    a[i]^=a[j];
                }
            }
        }
        for (int A:a)
            System.out.print((char)A+" ");
        return string;
    }
}

7　　只要把对应的ascii值+1即可

import java.util.Scanner;

public class Poseven {
    public static void main(String[] args) {
        //System.out.println(');
        Scanner input = new Scanner(System.in);
        System.out.print("请输入一串字符串:");
        String string = input.next();
        char []A = string.toCharArray();
        System.out.print("该字符串的密文为:");
        for (int i = 0; i < A.length; i++) {
            if ((A[i]>='a'&&A[i]<='z'||A[i]>='A'&&A[i]<='Z'))
                System.out.print((char)(A[i]+1));
            else
                System.out.print(A[i]);
        }
    }
}

8　　只要把对应的ascii值-1即可

import java.util.Scanner;

public class Poeight {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        System.out.print("请输入一串字符串：");
        String string = input.next();
        char []A = string.toCharArray();
        System.out.print("该字符串的明文为：");
        for (int i = 0; i < A.length; i++) {
            if ((A[i]>='a'&&A[i]<='z'||A[i]>='A'&&A[i]<='Z'))
                System.out.print((char)(A[i]-1));
            else
                System.out.print(A[i]);
        }
    }
}

9　　根据题目意思，在“ ”，“，”，“.”分割即可

public class Ponine {
    public static void main(String[] args) {
        String s = "no pains,no gains.";
        int count=0;
//        字符串变数组
        char[]A =s.toCharArray();
        char []B=new char[26];
        for (int AA:A){
            System.out.print((char)AA);
        }
        System.out.println();
      //  for (int i = 0; i <A.length ; i++) {
        int k=0;
           while (A[k]!='.') {
               if (A[k] == ' ' || A[k] == ',') {
                   k++;
                   System.out.println();
                   if (k == A.length)
                       k--;
               }
               System.out.print(A[k]);
               k++;
           }
     //   }
        System.out.println();
            int j=0;
        for (int i = 0; i <A.length ; i++)
            if (A[i] != ' ' && A[i] != ',' && A[i] != '.') {
                B[j] = A[i];
                j++;
                count++;
            }

    }
}

10　　熟悉cmd下执行java程序

 11　　冒泡法进行排序

 12、13  自行百度查看