Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.

Example 1:

Input: "babad"
Output: "bab"
Note: "aba" is also a valid answer.

Example 2:

Input: "cbbd"
Output: "bb" Solution 1:
Time: O(N^2)
class Solution:
def longestPalindrome(self, s: str) -> str:
if len(s) <= 1:
return s
self.res = ''
for i, char in enumerate(s):
self.helper(i, i, s) # odd
self.helper(i, i + 1, s) # even
return self.res def helper(self, start, end, s, even=False):
while start >= 0 and end < len(s):
if s[start] == s[end]:
if end - start >= len(self.res):
self.res = s[start: end + 1]
start -= 1
end += 1
else:
return
class Solution {
int start = 0;
int maxLen = 0;
public String longestPalindrome(String s) {
int len = s.length();
for (int i = 0; i < len; i++) {
helper(i, i, s);
helper(i, i + 1, s);
}
return s.substring(start, start + maxLen);
} private void helper(int low, int high, String s) {
while (low >= 0 && high < s.length() && s.charAt(low) == s.charAt(high)) {
low -= 1;
high += 1;
}
if (high - low - 1 > maxLen) {
maxLen = high - low - 1;
start = low + 1;
}
}
}

Solution 2:

Time: O(N^2)

class Solution {
public String longestPalindrome(String s) {
if (s == null || s.length() <= 1) {
return s;
}
boolean[][] isPalin = new boolean[s.length()][s.length()];
int max = 0;
String res = "";
for (int i = 1; i < s.length(); i++) {
// i == j for case of single char
for (int j = 0; j <= i; j++) {
if (s.charAt(i) == s.charAt(j) && (i - j <= 2 || isPalin[i - 1][j + 1])) {
isPalin[i][j] = true;
if (i - j + 1> max) {
max = i - j + 1;
// j is smaller than i
res = s.substring(j, i + 1);
}
}
}
}
return res;
}
}

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