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 两个pointer分别纪录当前window里两个character最后一次发生时的index,用start纪录window开始时的index。
从index 0开始遍历String:
如果当前的character在window内,update相应的pointer。
如果不在,比较两个character的pointer,去掉出现较早的那个character, 更新start=min(p1,p2)+1
时间复杂度是O(n), 空间复杂度是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 }