https://www.cnblogs.com/lightwindy/p/8526298.html
Given a string S, find the length of the longest substring T that contains at most two distinct characters.
For example,
Given S = "eceba",
T is "ece" which its length is 3.
For example,
Given S = "eceba",
T is "ece" which its length is 3.
给一个字符串,找出最多有两个不同字符的最长子串。还是滑动窗口法,定义一个HashMap或者Array来存储出现过的字符个数,左右指针初始位子为0,右指针向右扫,扫到新字符不同字符数加1,存入HashMap,扫到出现过的字符,HashMap对应字符数量加1。如果不同字符数大于2了,就把左指针位置的字符在HashMap中的数量减1,注意不一定是这一个,如果此时这个字符的数量大于0,说明还有这个字符存在,左指针加1继续右移,扫到的字符以后不会在窗口里了,要在HashMap中减1,如果这个字符减1后为0了,说明这个字符在窗口中没有了,此时不同字符数就可减1,左指针在右移1位指向下一个字符,统计此时窗口长度与最大长度比较,保留最大值。重复上面步骤,直到右指针扫完所有字符。最后返回最大长度。
public int lengthOfLongestSubstringTwoDistinct(String s) {        Map<Character, Integer> map = new HashMap<>();        int start = 0, end = 0;        int counter = 0, len = 0;        while(end < s.length()){            char cur = s.charAt(end);            if(!map.containsKey(cur)) map.put(cur, 0);            map.put(cur, map.get(cur) + 1);            if(map.get(cur) == 1) counter++;//new distinct char            while(counter > 2){//                char c2 = s.charAt(start);                map.put(c2, map.get(c2) - 1);                if(map.get(c2) == 0) counter--;                start++;            }            len = Math.max(end - start + 1, len);            end++;        }        return len;    }