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Leetcode-Longest Substring with At Most Two Distinct Characters.

时间:2014-12-13 06:07:09      阅读:137      评论:0      收藏:0      [点我收藏+]

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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.

Solution:

 1 public class Solution {
 2     public int lengthOfLongestSubstringTwoDistinct(String s) {
 3         //Two pointers: start, end. Maintain that start-end is a valid substring.
 4         //Record the number of the two chars in start-end.
 5         //Inc end, if the charAt(end) is a thrid char, then inc start, until the start-end is a valid substring again
 6         //record the length of the new string.
 7 
 8         Map<Character,Integer> charNum = new HashMap<Character,Integer>();
 9         int start = 0;
10         int end = 0;
11         int charType = 2;
12         int maxLen = 0;
13         while (end<s.length()){
14             char cur = s.charAt(end);
15             //if this char is in substring already, then increase its number
16             if (charNum.containsKey(cur))
17                 charNum.put(cur,charNum.get(cur)+1);
18             else {
19                 charNum.put(cur,1);
20                 if (charType>0) charType--;
21                 else {
22                    //We need eliminate another char in substring to maintain the feasibility of the substring.
23                    while (charNum.get(s.charAt(start))>1){
24                        charNum.put(s.charAt(start),charNum.get(s.charAt(start))-1);
25                        start++;
26                    }
27                    charNum.remove(s.charAt(start));
28                    start++;
29                 }
30             }
31 
32             if (maxLen<end-start+1) maxLen = end-start+1;
33             end++;
34         }
35 
36         return maxLen;        
37     }
38 }

 

Leetcode-Longest Substring with At Most Two Distinct Characters.

标签:style   blog   io   ar   color   os   sp   for   on   

原文地址:http://www.cnblogs.com/lishiblog/p/4160854.html

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