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给你一个字符串 s，找到 s 中最长的回文子串。
示例 1：
输入：s = "babad" 输出："bab" 解释："aba" 同样是符合题意的答案。
示例 2：
输入：s = "cbbd" 输出："bb"
示例 3：
输入：s = "a" 输出："a"
示例 4：
输入：s = "ac" 输出："a"
提示：
1 <= s.length <= 1000s 仅由数字和英文字母（大写和/或小写）组成动态规划法。
设 dp[i][j] 表示字符串 s[i..j] 是否为回文串。
j - i < 2，即字符串长度为 2 时，只要 s[i] == s[j]，dp[i][j] 就为 true。j - i >= 2，dp[i][j] = dp[i + 1][j - 1] && s[i] == s[j]。class Solution:
def longestPalindrome(self, s: str) -> str:
n = len(s)
dp = [[False] * n for _ in range(n)]
start, mx = 0, 1
for j in range(n):
for i in range(j + 1):
if j - i < 2:
dp[i][j] = s[i] == s[j]
else:
dp[i][j] = dp[i + 1][j - 1] and s[i] == s[j]
if dp[i][j] and mx < j - i + 1:
start, mx = i, j - i + 1
return s[start:start + mx]
class Solution {
public String longestPalindrome(String s) {
int n = s.length();
boolean[][] dp = new boolean[n][n];
int mx = 1, start = 0;
for (int j = 0; j < n; ++j) {
for (int i = 0; i <= j; ++i) {
if (j - i < 2) {
dp[i][j] = s.charAt(i) == s.charAt(j);
} else {
dp[i][j] = dp[i + 1][j - 1] && s.charAt(i) == s.charAt(j);
}
if (dp[i][j] && mx < j - i + 1) {
mx = j - i + 1;
start = i;
}
}
}
return s.substring(start, start + mx);
}
}
class Solution {
public:
string longestPalindrome(string s) {
int n = s.size();
vector<vector<bool>> dp(n, vector<bool>(n, false));
int start = 0, mx = 1;
for (int j = 0; j < n; ++j) {
for (int i = 0; i <= j; ++i) {
if (j - i < 2) {
dp[i][j] = s[i] == s[j];
} else {
dp[i][j] = dp[i + 1][j - 1] && s[i] == s[j];
}
if (dp[i][j] && mx < j - i + 1) {
start = i;
mx = j - i + 1;
}
}
}
return s.substr(start, mx);
}
};
func longestPalindrome(s string) string {
n := len(s)
dp := make([][]bool, n)
for i := 0; i < n; i++ {
dp[i] = make([]bool, n)
}
mx, start := 1, 0
for j := 0; j < n; j++ {
for i := 0; i <= j; i++ {
if j-i < 2 {
dp[i][j] = s[i] == s[j]
} else {
dp[i][j] = dp[i+1][j-1] && s[i] == s[j]
}
if dp[i][j] && mx < j-i+1 {
mx, start = j-i+1, i
}
}
}
return s[start : start+mx]
}
public class Solution{
public string LongestPalindrome(string s) {
int n = s.Length;
bool[,] dp = new bool[n, n];
int mx = 1, start = 0;
for (int j = 0; j < n; ++j)
{
for (int i = 0; i <= j; ++i)
{
if (j - i < 2)
{
dp[i, j] = s[i] == s[j];
}
else
{
dp[i, j] = dp[i + 1, j - 1] && s[i] == s[j];
}
if (dp[i, j] && mx < j - i + 1)
{
mx = j - i + 1;
start = i;
}
}
}
return s.Substring(start, mx);
}
}
import std/sequtils
proc longestPalindrome(s: string): string =
let n: int = s.len()
var
dp = newSeqWith[bool](n, newSeqWith[bool](n, false))
start: int = 0
mx: int = 1
for j in 0 ..< n:
for i in 0 .. j:
if j - i < 2:
dp[i][j] = s[i] == s[j]
else:
dp[i][j] = dp[i + 1][j - 1] and s[i] == s[j]
if dp[i][j] and mx < j - i + 1:
start = i
mx = j - i + 1
result = s[start ..< start+mx]
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