Question

我想在不使用内置函数的情况下为String回文制作程序。

以下是我到目前为止所尝试的代码：

public class Palindrom
    {
        private static Scanner in;

        public static void main(String[] args)
        {
            String s,str1,str2;
            Scanner scan =new Scanner (System.in);
            System.out.println("Enter the string");
            String s = in.nextLine();
            StringBuffer str1 = new StringBuffer();
            StringBuffer str2 = new StringBuffer();
            str1.reverse();
            System.out.println("orignal string="+str2);
            System.out.println("reveser string="+str1);
             if(String.valueOf(str1).compareTo(String.valueOf(str2))==0)

                System.out.println("palindrom");

             else
                 System.out.println("not palindrom");
        }
    }

这个程序运行不正常。我认为问题在in.nextLine和字符串缓冲区。

Answer 1

下面给出的是检查此处输入的代码是否为回文的最简单方法。

import java.util.Scanner;
public class Strng {
    public static void main(String[] args) {
    String r = ""; //To store the reverse
    Scanner sc = new Scanner(System.in);
    System.out.println("Enter the String");
    String s = sc.next(); // Entering the string

    for(int i= s.length() - 1;i>=0;i--) {
        r = r + s.charAt(i);
    }

    if(r.equals(s)) {
        System.out.println("Is a palindrome");
    }
    else {
        System.out.println("Not a palindrome");
    }
    }

}

Answer 2

public static void palindrome(String str){
        Map<Character,Integer> myMap = new HashMap<Character, Integer>();
        int characterCounter = 1;
        int index=-1;
        for(char ch : str.toCharArray()){
            index++;
            if(str.length()%2!=0 && index==Math.abs(str.length()/2)) continue;
            if(myMap.containsKey(ch))
                characterCounter = myMap.get(ch)+1;
            myMap.put(ch, characterCounter);
        }
        boolean flag = true;
        for(Character myMApVal: myMap.keySet()){
            if(myMap.get(myMApVal)%2!=0){
                flag=false;
                break;
            }               
        }
        if(!flag) System.out.println(str+" is not palindrome");
        else System.out.println(str+" is palindrome");      
}

Answer 3

package practice;

import java.util.Scanner;

public class String_Palindrome {
    public static void main(String[] args) {
//String s, str1, str2;
        Scanner scanner = new Scanner(System.in);
        System.out.println("Enter the string:");
        String s = scanner.nextLine();
        StringBuffer str1 = new StringBuffer(s);
        StringBuffer str2 = new StringBuffer(s);
        str1.reverse();
        //System.out.println("Original String is:" + str2);
        //System.out.println("Reverse String is:" + str1);
        if (String.valueOf(str1).compareTo(String.valueOf(str2)) == 0)
            System.out.println("Given String is a palindrome");
        else
            System.out.println("Given String is Not a palindrome");
    }
}

Answer 4

    import java.util.Scanner;

    public class Palindrom {

        static Scanner scanner = new Scanner(System.in);

        public static void main(String[] args) {

            System.out.println("Enter the string");
            String inputStr =scanner.next();
            System.out.println("Given String = "+inputStr);
            char [] charArray=inputStr.toCharArray();
            int strlength=(charArray.length)-1;
            boolean isPalindrom=true;

            for(int count=0; count<charArray.length && strlength >= 0; count++,strlength--){
                if(charArray[count]!=charArray[strlength]){
                    isPalindrom=false;
                    break;
                }      
            }

            if(isPalindrom){
                System.out.println("palindrom");
            }else{
                System.out.println("not palindrom");
            }

        }
    }

解释：

给定字符串以转换char数组。
取两点

字符串a [0]
的第一个开头
字符串a [length-1]
比较char数组中的每个char

比较开始到结束并结束循环开始。

Palindrom案例1 :(长度为奇数）

“女士”是Palindrom：

a [0] == a [4] m

a [1] == a [3] a

a [2] == a [2] d

a [3] == a [1] a

a [4] == a [0] m

Palindrom案2 :(长度均匀）

“malayalam”是Palindrom：

a [0] == a [8] m

a [1] == a [7] a

a [2] == a [6] l

a [3] == a [5] a

a [4] == a [4] y

a [5] == a [3] a

a [6] == a [2] l

a [7] == a [1] a

a [8] == a [0] m

Not Palindrom案例3：在每个字符中都不匹配（从开始到结束开始）。

Answer 5

import java.util.Scanner;

public class StringPalindrome {

public static void main(String[] args) {
    Scanner sc = null;
    String str1 = null;
    // create object for scanner class
    sc = new Scanner(System.in);
    if (sc != null) {
        System.out.println("Enter First String");
        str1 = sc.nextLine();
    }
    strPalindrom(str1);
}// main

static void strPalindrom(String str) {
    // converting string into array
    char ch[] = str.toCharArray();
    // check string is Palindrom or not
    int count = ch.length - 1;
    for (int i = 0; i < ch.length; i++, count--) {
        if (ch[i] != ch[count]) {
            System.out.println("String is not Palindrom");
            break;
        } else {
            if (i == count) {
                System.out.println("String is Palindrom");
            }
        }
    } // for
}// strPalindrom method
}// class

Answer 6

我在上面看到了所有程序，但对初学者来说有点难以理解。

非常简单的初学者课程（检查String Palindrome）

public class Palindrome{
public static void main(String[] args) {
    String s = "malayalam"; // enter here your requred String to check Palindrom
    String rev = "";
    for (int i = s.length() - 1; i >= 0; i--) {
        rev = rev + s.charAt(i);
    }
    if (rev.equals(s)) {
        System.out.println("Palindrom");
    } else {
        System.out.println("Not Palindrom");
    }
  }

}

Answer 7

import java.util.*;
class Test{

    public static void main(String... as){
        int check=0;
        Scanner s= new Scanner(System.in);
        String str = s.nextLine();
        int l=str.length();
        for (int i=0;i<l;i++){
            if(str.charAt(i)!=str.charAt(l-1-i))
                check++;
        }
        if(check==0)
            System.out.println("yes it's palindrome");
        else
            System.out.println("no it's not.");

}
}

Answer 8

public class Palindrom {

        public static void main(String[] args) {

        String str= "dad";
        char[] rev=str.toCharArray();
        int i=0;
        int j= rev.length-1;
        while(i<j){
            char temp =rev[i];
            rev[i] = rev[j];
            rev[j] = temp;
            i++;
            j--;
        }
        System.out.println(rev);
        System.out.println();
        if(str.equals(String.valueOf(rev))){
            System.out.println("Yes! It is palindrom");
        }else{
            System.out.println("No! It is not palindrom");
        }
    }

}

不使用内置函数的字符串回文

8 个答案: