# 5 Longest Palindromic Substring

### Problem:

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.

### Thoughts:

Trivial way is to calculate each substring is palindrome, using O(n^2) time and space. Could be optimized using only O(1) space. Idea is to return the longest palindromic substring that is center by (i,i) or (i, i + 1).

### Solution:

Trivial way:

``````public class Solution {
public String longestPalindrome(String s) {
boolean[][] isPa = new boolean[s.length()][s.length()];
calIsPa(isPa, s);
String longest = "";
int max = 0;
for (int i = 0; i < s.length(); i ++){
for (int j = i; j < s.length(); j ++){
if (isPa[i][j] == true && (j - i + 1) > max){
max = j - i + 1;
longest = s.substring(i, j + 1);
}
}
}
return longest;
}
private void calIsPa(boolean[][] isPa, String s){
for (int i = 0; i < s.length(); i ++){
isPa[i][i] = true;
}
for (int i = 0; i < s.length() - 1; i ++){
if (s.charAt(i) == s.charAt(i+1)){
isPa[i][i+1] = true;
}
}
for (int l = 3; l <= s.length(); l ++ ){
for (int i = 0; i < s.length() - l + 1; i ++){
if (s.charAt(i) == s.charAt(i + l - 1) && isPa[i+1][i + l -2]){
isPa[i][i+ l - 1] = true;
}
}
}
}

}
``````

The better way:

``````public class Solution {
public String longestPalindrome(String s) {
String longest = "";
for (int i = 0; i < s.length(); i ++){
String tmp = "";
tmp = paCenterBy(s, i, i);
if (tmp.length() > longest.length()){
longest = tmp;
}
tmp = paCenterBy(s, i, i + 1);
if (tmp.length() > longest.length()){
longest = tmp;
}
}
return longest;
}
private String paCenterBy(String s, int cleft, int cright){
String result = "";
while (cleft >=0 && cright < s.length() && s.charAt(cleft) == s.charAt(cright)){
cleft --;
cright ++;
}
return s.substring(cleft + 1, cright);
}
}
``````