题解 | 穷哈哈~

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner scanner=new Scanner(System.in);
		int n=scanner.nextInt();
		String string=scanner.next();
		int flag=0;
		if(string.contains("a")) {
			flag++;
		}
		if(string.contains("h")){
			flag++;
		}
		if(flag==0) {
			System.out.println(0);
		}else if(flag==1){
			System.out.println(1);
		}else if(flag==2) {
			int count=1;
			int max=-1;
			for (int i = 0; i < n-1; i++) {
				char c1=string.charAt(i);
				char c2=string.charAt(i+1);
				if(c1=='a'&&c2=='h'||c1=='h'&&c2=='a') {
					count++;
				}else {
					if(count>max)max=count;
					count=1;
				}
			}
            if(count>max)max=count;
			System.out.println(max);
		}
		
		

	}

}

学会了一种新的解法，就是一开始就判断是否含有a或者h，根据含有情况，分别输出0，1，如果都有，那么在进行交替数量统计