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原题链接在这里：https://leetcode.com/problems/longest-substring-with-at-most-two-distinct-characters/
与Longest Substring Without Repeating Characters类似。
维护一个窗口[walker, runner]. 同时维护一个HashMap<Character,Integer>, key是字符, value是字符出现次数。要保证hm.size()<=2.
在runner<s.length()的条件下，若是hm已经包含s.charAt(runner), 对应数量加一即可。
若是不包含s.charAt(runner), 此时要看hm.size()是否小于2, 若是小于2, 插入新的数据即可。若是已经等于2了，说明不能插入了，此时就要开始移动walker.
从hm中减掉walker对应的char的数量，知道hm.size()不再等于2即可。
Time Complexity: O(n), 窗口只移动一遍. Space: O(1). HashMap的大小控制在2.
AC Java:
1 public class Solution { 2 public int lengthOfLongestSubstringTwoDistinct(String s) { 3 if(s == null || s.length() == 0){ 4 return 0; 5 } 6 7 int res = 0; 8 HashMap<Character,Integer> hm = new HashMap<Character,Integer>(); 9 10 int walker = 0; 11 int runner = 0; 12 while(runner < s.length()){ 13 if(hm.containsKey(s.charAt(runner))){ 14 hm.put(s.charAt(runner), hm.get(s.charAt(runner))+1); 15 runner++; 16 }else if(hm.size() < 2){ 17 hm.put(s.charAt(runner), 1); 18 runner++; 19 }else{ 20 while(hm.size() == 2){ 21 hm.put(s.charAt(walker),hm.get(s.charAt(walker))-1); 22 if(hm.get(s.charAt(walker)) == 0){ 23 hm.remove(s.charAt(walker)); 24 } 25 walker++; 26 } 27 } 28 res = Math.max(res,runner-walker); 29 } 30 return res; 31 } 32 }
LeetCode Longest Substring with At Most Two Distinct Characters
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原文地址：http://www.cnblogs.com/Dylan-Java-NYC/p/5182617.html
