Longest Palindromic Substring

Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring.

 1 public class Solution {
 2     String longestOne = null;
 3     int lengthOfLongest = 0;
 4     public String longestPalindrome(String s) {
 5         // Start typing your Java solution below
 6         // DO NOT write main() function
 7         longestOne = null;
 8         lengthOfLongest = 0;
 9         if(s.length() == 0){
10             return longestOne;
11         }
12         if(s.length() == 1){
13             return s;
14         }
15         for(int i = 0; i < s.length() - 1; i ++){
16             if(s.charAt(i) == s.charAt(i+1)){
17                 if((i == 0 || i == s.length() - 2) && lengthOfLongest < 2){
18                         lengthOfLongest = 2;
19                         longestOne = s.substring(i,i+2);
20                 }
21                 else{getLongestPalindrome(i - 1, i + 2, s);}
22             }
23         }
24         for(int i = 1; i < s.length() - 1; i ++){
25             getLongestPalindrome(i - 1, i + 1, s);
26         }
27         return longestOne;
28     }
29     
30     public void getLongestPalindrome(int left, int right, String s){
31         if(left < 0 || right >= s.length()) return;
32         int j = right;
33         for(int i = left; i > -1 ; i --){
34             if(j < s.length() && s.charAt(i) == s.charAt(j)){
35                 if(j-i+1 >= lengthOfLongest){
36                     lengthOfLongest = j-i+1;
37                     longestOne = s.substring(i,j+1);
38                 }
39             }
40             else{
41                 return;
42             }
43             j ++;
44         }
45     }
46 }

第二遍：

可以将长度为奇数的substring 和偶数substring 归并到一起。即使用两倍长度进行循环。如果是点则以为中心，如果是中间，则以两边的点为中心。减少很多test case。

 1 public class Solution {
 2     public String longestPalindrome(String s) {
 3         // Note: The Solution object is instantiated only once and is reused by each test case.
 4         String longestOne = null;
 5         int max = 0;
 6         int len = s.length();
 7         for(int i = 0; i <= (len - 1)* 2; i ++){
 8             int left = (i % 2 == 1 ? i / 2 : i / 2 - 1);
 9             int right = i / 2 + 1;
10             while(left > -1 && right < len && s.charAt(left) == s.charAt(right)){
11                 left --;
12                 right ++;
13             }
14             if(right - left - 1 > max){
15                 max = right - left - 1;
16                 longestOne = s.substring(left + 1, right);
17             }
18         }
19         return longestOne;
20     }
21 }

第三遍：

 1 public class Solution {
 2     public String longestPalindrome(String s) {
 3         if(s == null || s.length() == 0) return null;
 4         int len = s.length();
 5         int max = 0;
 6         int max_left = 0, max_right = 0;
 7         for(int i = 0; i < len * 2 - 1; i ++){
 8             int left  = i / 2;
 9             int right = (i + 1) / 2;
10             while(left > -1 && right < len && s.charAt(left) == s.charAt(right)){
11                 left --;
12                 right ++;
13             }
14             if(right - left > max){
15                 max = right - left;
16                 max_left = left + 1;
17                 max_right = right;
18             }
19         }
20         return s.substring(max_left,max_right);
21     }
22 }