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++) {
10 int len = j - i;
11 String curr = s.substring(i, j + 1);
12 if (isPalindrome(curr)) {
13 if (len > maxPalinLength) {
14 longestPalindrome = curr;
15 maxPalinLength = len;
16 }
17 }
18 }
19 }
20
21 return longestPalindrome;
22 }
23
24 public boolean isPalindrome(String s) {
25
26 for ( int i = 0; i < s.length() - 1; i++) {
27 if (s.charAt(i) != s.charAt(s.length() - 1 - i)) {
28 return false;
29 }
30 }
31
32 return true;
33 }
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++) {
10 int len = j - i;
11 String curr = s.substring(i, j + 1);
12 if (isPalindrome(curr)) {
13 if (len > maxPalinLength) {
14 longestPalindrome = curr;
15 maxPalinLength = len;
16 }
17 }
18 }
19 }
20
21 return longestPalindrome;
22 }
23
24 public boolean isPalindrome(String s) {
25
26 for ( int i = 0; i < s.length() - 1; i++) {
27 if (s.charAt(i) != s.charAt(s.length() - 1 - i)) {
28 return false;
29 }
30 }
31
32 return true;
33 }
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
10 if (tmp.length() > longest.length())
11 longest = tmp;
12
13 // get longest palindrome with center of i, i+1
14 tmp = helper(s, i, i + 1);
15 if (tmp.length() > longest.length())
16 longest = tmp;
17 }
18
19 return longest;
20 }
21
22 // Given a center, either one letter or two letter,
23 // Find longest palindrome
24 public String helper(String s, int begin, int end) {
25 while (begin >= 0 && end <= s.length() - 1 && s.charAt(begin) == s.charAt(end)) {
26 begin--;
27 end++;
28 }
29 return s.substring(begin + 1, end);
30 }
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
10 if (tmp.length() > longest.length())
11 longest = tmp;
12
13 // get longest palindrome with center of i, i+1
14 tmp = helper(s, i, i + 1);
15 if (tmp.length() > longest.length())
16 longest = tmp;
17 }
18
19 return longest;
20 }
21
22 // Given a center, either one letter or two letter,
23 // Find longest palindrome
24 public String helper(String s, int begin, int end) {
25 while (begin >= 0 && end <= s.length() - 1 && s.charAt(begin) == s.charAt(end)) {
26 begin--;
27 end++;
28 }
29 return s.substring(begin + 1, end);
30 }
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