Pattern 3 (Inferred): 7. Longest Palindromic Substring

Description: Given a string s, return the longest palindromic substring in s.

Java

public String longestPalindrome(String s) {
if (s == null || s.length() == 0) return "";
int start = 0, maxLength = 1;
for (int i = 0; i < s.length(); i++) {
int len1 = expandAroundCenter(s, i, i);
int len2 = expandAroundCenter(s, i, i + 1);
int len = Math.max(len1, len2);
if (len > maxLength)
{ maxLength = len;
start = i - (maxLength - 1) / 2;
}
}
return s.substring(start, start + maxLength);
}
private int expandAroundCenter(String s, int left, int right) {

while (left >= 0 && right < s.length() && s.charAt(left) == s.charAt(right)) { left–; right++; } return right - left - 1; } C++: cpp string longestPalindrome(string s) { int start = 0, maxLength = 1; for (int i = 0; i < s.size(); i++) { int len1 = expandAroundCenter(s, i, i); int len2 = expandAroundCenter(s, i, i + 1); int len = max(len1, len2); if (len > maxLength) { maxLength = len; start = i - (maxLength - 1) / 2; } } return s.substr(start, maxLength); } int expandAroundCenter(string s, int left, int right) { while (left >= 0 && right < s.size() && s[left] == s[right]) { left–; right++; } return right - left - 1; }

Python

def longestPalindrome(s):
def expandAroundCenter(s, left, right):
while left >= 0 and right < len(s) and s[left] == s[right]:

left -= 1 right += 1 return right - left - 1 start, maxLength = 0, 1 for i in range(len(s)): len1 = expandAroundCenter(s, i, i) len2 = expandAroundCenter(s, i, i + 1) length = max(len1, len2) if length > maxLength: maxLength = length start = i - (maxLength - 1) // 2 return s[start:start + maxLength]