Leetcode: 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.

Analysis: 想到了Naive的方法，时间复杂度为O(N^3)，知道肯定不是最优的方法，一时也想不出别的方法，于是看了网上的做法，最后采纳了这个人的方法，感觉他讲的比较全面：http://www.programcreek.com/2013/12/leetcode-solution-of-longest-palindromic-substring-java/

1. Naive Approach:  The time complexity is O(n^3). If this is submitted to LeetCode onlinejudge, "Time Limit Exceeded".

2. Dynamic Programming: 

Let s be the input string, i and j are two indices of the string.

Define a 2-dimension array "table" and let table[i][j] denote whether substring from i to j is palindrome.

Start condition:

table[i][i] == 1;

table[i][i+1] == 1  => s.charAt(i) == s.charAt(i+1) 

Changing condition:

table[i][j] == 1 => table[i+1][j-1] == 1 && s.charAt(i) == s.charAt(j)

Time O(n^2) Space O(n^2)

以下代码参考了Code Ganker的，这种方法使用两层循环，时间复杂度是O(n^2)

 1 public String longestPalindrome(String s) {

 2     if(s == null || s.length()==0)

 3         return "";

 4     boolean[][] palin = new boolean[s.length()][s.length()];

 5     String res = "";

 6     int maxLen = 0;

 7     for(int i=s.length()-1;i>=0;i--)

 8     {

 9         for(int j=i;j<s.length();j++)

10         {

11             if(s.charAt(i)==s.charAt(j) && (j-i<=2 || palin[i+1][j-1]))

12             {

13                 palin[i][j] = true;

14                 if(maxLen<j-i+1)

15                 {

16                     maxLen=j-i+1;

17                     res = s.substring(i,j+1);

18                 }

19             }

20         }

21     }

22     return res;

23 }

3. Smart Algorithm: Time O(n^2), Space O(1)

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）往两边同时进行扫描，直到不是回文串为止。在这个过程中同时寻找最长Palindrome串

 1 public class Solution {

 2     public String longestPalindrome(String s) {

 3         if (s == null || s.length() == 0) return null;

 4         if (s.length() == 1) return s;

 5         String longest = "";

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

 7             String temp = helper(s, i, i);

 8             if (temp.length() > longest.length()) {

 9                 longest = temp;

10             }

11             temp = helper(s, i, i + 1);

12             if (temp.length() > longest.length()) {

13                 longest = temp;

14             }

15         }

16         return longest;

17     }

18     

19     public String helper(String s, int start, int end) {

20         while (start >=0 && end < s.length() && s.charAt(start) == s.charAt(end)) {

21             start--;

22             end++;

23         }

24         start++;

25         end--;

26         return s.substring(start, end + 1);

27     }

28 }