题解 | #包含不超过两种字符的最长子串#

import java.util.Scanner;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.StreamTokenizer;
import java.util.BitSet;
// 注意类名必须为 Main, 不要有任何 package xxx 信息
public class Main {
    public static void main(String[] args) throws Exception {
        StreamTokenizer st = new StreamTokenizer(new BufferedReader(
                    new InputStreamReader(System.in)));
        st.nextToken();
        String s = st.sval;
        int len = 0;
        int left = 0, right = 0;
        int pre = 0;
	  	//保存不重复字符的个数
        BitSet bs = new BitSet(3);
        while (right < s.length()) {
            char c = s.charAt(right);
            int size = bs.cardinality();
            bs.set(c);
		  //如果遇到第二个字符，存储当前位置
            if (size == 1 && bs.cardinality() == 2) {
                pre = right;
            }
		  	//如果字符个数小于等于两个，遍历下一个字符
            if (bs.cardinality() <= 2) {
                len = Math.max(len, right - left + 1);
                right++;
            } else {
			  	//如果字符数大于两个，从上一次存储的位置开始重新遍历
                left = pre;
                right = left;
                bs.clear();
            }
        }
        System.out.println(len);
    }
}