class Solution:
def longestPalindrome(self, s: str) -> str:
n = len(s)
if n == 0:
return ""
res = s[0]
def extend(i, j, s):
while(i >= 0 and j < len(s) and s[i] == s[j]):
i -= 1
j += 1
return s[i + 1:j]
for i in range(n - 1):
e1 = extend(i, i, s)
e2 = extend(i, i + 1, s)
if max(len(e1), len(e2)) > len(res):
res = e1 if len(e1) > len(e2) else e2
return res
JavaScript Code：
/*
* @lc app=leetcode id=5 lang=javascript
*
* [5] Longest Palindromic Substring
*/
/**
* @param {string} s
* @return {string}
*/
var longestPalindrome = function (s) {
// babad
// tag : dp
if (!s || s.length === 0) return "";
let res = s[0];
const dp = [];
// 倒着遍历简化操作， 这么做的原因是dp[i][..]依赖于dp[i + 1][..]
for (let i = s.length - 1; i >= 0; i--) {
dp[i] = [];
for (let j = i; j < s.length; j++) {
if (j - i === 0) dp[i][j] = true;
// specail case 1
else if (j - i === 1 && s[i] === s[j]) dp[i][j] = true;
// specail case 2
else if (s[i] === s[j] && dp[i + 1][j - 1]) {
// state transition
dp[i][j] = true;
}
if (dp[i][j] && j - i + 1 > res.length) {
// update res
res = s.slice(i, j + 1);
}
}
}
return res;
};
CPP Code:
class Solution {
private:
int expand(string &s, int L, int R) {
while (L >= 0 && R < s.size() && s[L] == s[R]) {
--L;
++R;
}
return R - L - 1;
}
public:
string longestPalindrome(string s) {
if (s.empty()) return s;
int start = 0, maxLen = 0;
for (int i = 0; i < s.size(); ++i) {
int len1 = expand(s, i, i);
int len2 = expand(s, i, i + 1);
int len = max(len1, len2);
if (len > maxLen) {
start = i - (len - 1) / 2;
maxLen = len;
}
}
return s.substr(start, maxLen);
}
};
