Given a string, find the length of the longest substring T that contains at most 2 distinct characters.

For example, Given s =?“eceba”,

T is "ece" which its length is 3.

用p1 & p2 兩個pointer分別紀錄當前window里兩個character最后一次發生時的index，用start紀錄window開始時的index。
從index 0開始遍歷String:

如果當前的character在window內，update相應的pointer。

如果不在，比較兩個character的pointer，去掉出現較早的那個character, 更新start=min(p1,p2)+1

時間復雜度是O(n), 空間復雜度是O(1):

?

1 public class Solution { 2 public int lengthOfLongestSubstringTwoDistinct(String s) { 3 int result = 0; 4 int first = -1, second = -1; 5 int win_start = 0; 6 for(int i = 0; i < s.length(); i ++){ 7 if(first < 0 || s.charAt(first) == s.charAt(i)) first = i; 8 else if(second < 0 || s.charAt(second) == s.charAt(i)) second = i; 9 else{ 10 int min = first < second ? first : second; 11 win_start = min + 1; 12 if(first == min) first = i; 13 else second = i; 14 } 15 result = Math.max(result, i - win_start + 1); 16 } 17 return result; 18 } 19 }

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轉載于:https://www.cnblogs.com/reynold-lei/p/4247706.html

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