[Link] https://leetcode.com/problems/longest-palindromic-substring/
Approach
Every palindrome has either a character at its center (an odd-length palindrome) or a gap between two characters at its center (an even-length palindrome). For each index, expand outward from both kinds of centers while the characters match. Each expansion checks all palindromes with that center.
Keep the best answer as a half-open interval [bestStart, bestEnd). When a longer palindrome is found, replace the interval; otherwise leave it unchanged so an existing maximum-length answer is preserved. An empty string has no centers and returns the empty substring.
Java
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class Solution {
public String longestPalindrome(String s) {
int bestStart = 0;
int bestEnd = 0;
for (int center = 0; center < s.length(); center++) {
int left = center;
int right = center;
while (left >= 0 && right < s.length()
&& s.charAt(left) == s.charAt(right)) {
left--;
right++;
}
if (right - left - 1 > bestEnd - bestStart) {
bestStart = left + 1;
bestEnd = right;
}
left = center;
right = center + 1;
while (left >= 0 && right < s.length()
&& s.charAt(left) == s.charAt(right)) {
left--;
right++;
}
if (right - left - 1 > bestEnd - bestStart) {
bestStart = left + 1;
bestEnd = right;
}
}
return s.substring(bestStart, bestEnd);
}
}
For each of the N positions, two expansions can each inspect up to N characters, so the time complexity is O(N²). The indices use only constant extra space, giving O(1) extra space.