From 40099647a18b208712d05cb313620e1738105b69 Mon Sep 17 00:00:00 2001
From: Harvey <harvey0379@163.com>
Date: Sat, 24 Aug 2024 15:12:58 +0800
Subject: [PATCH] [python]longest-palindromic-substring

---
 .../00005.longest-palindromic-substring/5.py  | 37 +++++++++++++++++++
 1 file changed, 37 insertions(+)
 create mode 100644 src/leetcode/00005.longest-palindromic-substring/5.py

diff --git a/src/leetcode/00005.longest-palindromic-substring/5.py b/src/leetcode/00005.longest-palindromic-substring/5.py
new file mode 100644
index 0000000..32151fb
--- /dev/null
+++ b/src/leetcode/00005.longest-palindromic-substring/5.py
@@ -0,0 +1,37 @@
+class Solution:
+    def longestPalindrome(self, s: str) -> str:
+        n = len(s)
+        if n < 2:
+            return s
+
+        max_len = 1
+        begin = 0
+        # dp[i][j] 表示 s[i..j] 是否是回文串
+        dp = [[False] * n for _ in range(n)]
+        for i in range(n):
+            dp[i][i] = True
+
+        # 递推开始
+        # 先枚举子串长度
+        for L in range(2, n + 1):
+            # 枚举左边界，左边界的上限设置可以宽松一些
+            for i in range(n):
+                # 由 L 和 i 可以确定右边界，即 j - i + 1 = L 得
+                j = L + i - 1
+                # 如果右边界越界，就可以退出当前循环
+                if j >= n:
+                    break
+
+                if s[i] != s[j]:
+                    dp[i][j] = False
+                else:
+                    if j - i < 3:
+                        dp[i][j] = True
+                    else:
+                        dp[i][j] = dp[i + 1][j - 1]
+
+                # 只要 dp[i][L] == true 成立，就表示子串 s[i..L] 是回文，此时记录回文长度和起始位置
+                if dp[i][j] and j - i + 1 > max_len:
+                    max_len = j - i + 1
+                    begin = i
+        return s[begin : begin + max_len]