题目

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.

题解

第一种方法就是挨个检查,维护全局最长,时间复杂度为O(n3),会TLE。

代码如下:

 1 public String longestPalindrome(String s) {

 2  

 3     int maxPalinLength = 0;

 4     String longestPalindrome = null;

 5     int length = s.length();

 6  

 7     // check all possible sub strings

 8     for (int i = 0; i < length; i++) {

 9         for (int j = i + 1; j < length; j++) {

             int len = j - i;

             String curr = s.substring(i, j + 1);

             if (isPalindrome(curr)) {

                 if (len > maxPalinLength) {

                     longestPalindrome = curr;

                     maxPalinLength = len;

                 }

             }

         }

     }

  

     return longestPalindrome;

 }

  

 public boolean isPalindrome(String s) {

  

     for (int i = 0; i < s.length() - 1; i++) {

         if (s.charAt(i) != s.charAt(s.length() - 1 - i)) {

             return false;

         }

     }

  

     return true;

 }

第二种方法“是对于每个子串的中心(可以是一个字符,或者是两个字符的间隙,比如串abc,中心可以是a,b,c,或者是ab的间隙,bc的间隙,例如aba是回文,abba也是回文,这两种情况要分情况考虑)往两边同时进
行扫描,直到不是回文串为止。假设字符串的长度为n,那么中心的个数为2*n-1(字符作为中心有n个,间隙有n-1个)。对于每个中心往两边扫描的复杂
度为O(n),所以时间复杂度为O((2*n-1)*n)=O(n^2),空间复杂度为O(1)。”引自Code ganker(http://codeganker.blogspot.com/2014/02/longest-palindromic-substring-leetcode.html)

代码如下:

 1     public String longestPalindrome(String s) {
 2         if (s.isEmpty()||s==null||s.length() == 1)
 3             return s;
 4      
 5         String longest = s.substring(0, 1);
 6         for (int i = 0; i < s.length(); i++) {
 7             // get longest palindrome with center of i
 8             String tmp = helper(s, i, i);
 9             
             if (tmp.length() > longest.length())
                 longest = tmp;
      
             // get longest palindrome with center of i, i+1
             tmp = helper(s, i, i + 1);
             if (tmp.length() > longest.length())
                 longest = tmp;
         }
      
         return longest;
     }
      
     // Given a center, either one letter or two letter, 
     // Find longest palindrome
     public String helper(String s, int begin, int end) {
         while (begin >= 0 && end <= s.length() - 1 && s.charAt(begin) == s.charAt(end)) {
             begin--;
             end++;
         }
         return s.substring(begin + 1, end);
     }

Reference:

http://www.programcreek.com/2013/12/leetcode-solution-of-longest-palindromic-substring-java/

http://codeganker.blogspot.com/2014/02/longest-palindromic-substring-leetcode.html

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