JAVA palindromic sequence is determined recursively

Palindromic sequence is determined recursively

Ideas:

① First, if the empty string, a character , it is determined palindromic sequence

② not satisfied ①, then we need to string a recursive judgment, we thought to use the first character and the last character comparison, if the same is the second and penultimate compare, so the cycle , there will be an end

③, two things occur at the end, the first case: two different positions of characters are compared, the second case is: the character of the two identical positions compared. So we need to find the middle value , we pass sub parity manner, calculates the final left subscript numbers to be compared, and if it is odd, the left subscript of: len / 2 ; if for the even-numbered subscripts, is left : len / 2 ;

With recursive end condition ④, then the recursion conditions are: when the first character is the same, it is recursive: index +1 left, right subscript -1 if not met, then exit.

Code:

package com.java1;
import java.util.Scanner;
public class test {
    static String s=new String();
    static int len=0;
    static int k;
public static void  f(int i,int j)
{
    if(i==k&&s.charAt(i)==s.charAt(j))//判断结束的条件
    {
        System.out.println("YES");
        return;
    }
    
    if(s.charAt(i)==s.charAt(j))//If yes the next step is performed recursively 
        F (I +. 1,. 1-J );
     the else // not satisfied, the output 
    { 
        System.out.println ( "NO" );
         return ; 
    } 
} 
    public  static  void main (String [ ] args) {
         // the TODO Auto-Generated Method Stub 
    Scanner in = new new Scanner (the System.in); 
    S = in.nextLine ();
     int K = 0 ; 
    len = s.length (); // Get string length 
    IF (len% 2 == 0) // to find the intermediate value by the number of parity 
        K = len / 2-1; //Find the middle subscript 
    the else 
        K = len / 2; // find the middle subscript 
    IF (len len == == 0 ||. 1) // If an empty string or only a character 
        System.out.println ( "OK" );
     the else 
    { 
        F ( 0,. 1-len); // recursive computation 
    } 
    } 
}