# Share my naive O(n^2) Java solution (beats 82.11%)

• -- the idea is simple: just expand from one center to see how far it can go

-- iterate through all centers

-- note that for each center/expansion, the length of palindromic substring can be odd or even.

public class Solution {
public String longestPalindrome(String s) {
int from = 0, to = 0;
for (int i = 0; i < s.length(); ++i) {
int[] oddLenResult = expandFromCenter(s, i - 1, i + 1);
if (oddLenResult[1] - oddLenResult[0] > to - from) {
from = oddLenResult[0];
to = oddLenResult[1];
}
int[] evenLenResult = expandFromCenter(s, i, i + 1);
if (evenLenResult[1] - evenLenResult[0] > to - from) {
from = evenLenResult[0];
to = evenLenResult[1];
}
}
return s.substring(from, to);
}

private int[] expandFromCenter(String s, int left, int right) {
int p = left, q = right;
while (p >= 0 && q < s.length() && s.charAt(p) == s.charAt(q)) {
--p;
++q;
}
return new int[] {p + 1, q};  // s.substring(p + 1, q) is the current longest palindromic substring
}
}

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