题解 | 穷哈哈~

import java.util.Scanner;

/**
 * @author supermejane
 * @date 2025/10/3
 * @description 牛客算法学习篇-算法入门   BGN5穷哈哈~
 */
public class Bgn5 {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int len = in.nextInt();
        in.nextLine();
        String s = in.nextLine();
		
	  
	  	//解法一：直接单个指针遍历就行
        int i = 0, result = Integer.MIN_VALUE, subLen = 0;
        while (i < len) {
            if (illegal(s, i)) {
                subLen++;
            } else {
			  	//非字串部分，直接结束字串
                result = Math.max(subLen, result);
                subLen = s.charAt(i) == 'a' || s.charAt(i) == 'h' ? 1 : 0;
			}
            i++;
        }
        result = Math.max(result, subLen);
        System.out.println(result);
    }
	
  	//字串部分
    public static boolean illegal(String s, int i) {
        return (s.charAt(i) == 'a' && (i == 0 || s.charAt(i - 1) == 'h')) || (s.charAt(i) == 'h' && (i == 0 || s.charAt(i - 1) == 'a'));
    }

  
  
  
    //下面是解法二：双指针
    public static int subLength(String s) {
        //p1->s, p2->sub字串
        int p1 = 0, p2 = 0 , result = 0, subLen = 0;

        while (p1 < s.length()) {
            if (illegal2(s, p1)) {
                p2 = p1 + 1;
                subLen = 1;
                while (p2 < s.length() && ((s.charAt(p2) == 'a' && s.charAt(p2 - 1) == 'h') || (s.charAt(p2) == 'h' && s.charAt(p2 - 1) == 'a'))) {
                    subLen++;
                    p2++;
                }
                p1 = p2;
                result = Math.max(result, subLen);
            } else {
                p1++;
            }
        }
        return result;
    }

    //子串开始
    public static boolean illegal2(String s, int i) {
        if (s.charAt(i) == 'a') {
            return i == 0 || s.charAt(i) != 'h';
        } else if (s.charAt(i) == 'h') {
            return i == 0 || s.charAt(i) != 'a';
        }
        return false;
    } 
}