算法: 最长回文子字符串5. Longest Palindromic Substring

5. Longest Palindromic Substring

Given a string s, return the longest
palindromic

substring
in s.

Example 1:

Input: s = "babad"
Output: "bab"
Explanation: "aba" is also a valid answer.

Example 2:

Input: s = "cbbd"
Output: "bb"

Constraints:

• 1 <= s.length <= 1000
• s consist of only digits and English letters.

1. 两个指针解法，注意中间位置是单个数还是双数

class Solution:
    def longestPalindrome(self, s: str) -> str:
        def expand(i, j):
            left = i
            right = j

            while left >= 0 and right < len(s) and s[left] == s[right]:
                left -= 1
                right += 1
            
            return right - left - 1
        
        ans = [0, 0]
        for i in range(len(s)):
            odd_len = expand(i ,i)
            if odd_len > ans[1] - ans[0] + 1:
                dist = odd_len // 2
                ans = [i - dist, i + dist]
            
            even_len = expand(i, i+1)
            if even_len > ans[1] - ans[0] + 1:
                dist = (even_len // 2) - 1
                ans = [i - dist, i + 1 + dist]
        
        i, j = ans
        return s[i: j+1]

2. 动态规划解法

1. 动态规划公式 s[i] == s[j] and dp[i+1][j-1]， 如果两头相等，并且子集dp[i+1][j-1] 已经证实过，就扩大回文到[i, j]
2. 必须从小到大，所以就要设置步长diff。并且优先处理单步的问题s[i] == s[i+1]

class Solution:
    def longestPalindrome(self, s: str) -> str:
        n = len(s)
        dp = [[False] * n for _ in range(n)]
        ans = [0, 0]

        for i in range(n):
            dp[i][i] = True
        
        for i in range(n - 1):
            if s[i] == s[i+1]:
                dp[i][i+1] = True
                ans = [i, i+1]
        
        for diff in range(2, n):
            for i in range(n - diff):
                j = i + diff
                if s[i] == s[j] and dp[i+1][j-1]:
                    dp[i][j] = True
                    ans = [i, j]
        
        i, j = ans
        return s[i: j+1]