Leedcode daily punch 4.12

 Longest Palindromic Substring - LeetCode (leetcode-cn.com)

Given a string s, find the longest palindrome substring in s.

Example 1:

Input: s = "babad"
Output: "bab"
Explanation: "aba" is also the answer that meets the meaning of the question.
Example 2:

Input: s = "cbbd"
Output: "bb"
 

hint:

1 <= s.length <= 1000
s only consists of numbers and English letters

Source: LeetCode

 dynamic programming

Idea: Define dp[i][j] as whether the string starting from the i-th character of the string to the end of the j-th character is a palindrome

(i<=j)

Substring s[l,r], if s[l+1:r-1] is a palindrome, then just check whether s[l] and s[r] are equal

即dp[l][r]=dp[l+1][r-1] and s[l]==s[r]

In a special case, if l+1 is equal to r-1, it means that there is only one element between the two endpoints,

dp[l][r]=True if s[l]==s[r] else False, in fact, the above case can also be incorporated, because an element is a palindrome by itself

If l+1>r-1, it means that there is no element between the two endpoints, then it is only necessary to directly judge whether s[i] and s[r] are equal.

In the above explanation process, we all emphasized the two endpoints, but because i<=j, when i=j, the two points overlap, and the assignment can be done directly.

Summary: Define the meaning of the equation, find out the state transition equation, and determine the boundary conditions based on the state transition equation.

class Solution:
    def longestPalindrome(self, s: str) -> str:
        n=len(s)
        dp=[[False]*n for i in range(n)]
        dp[n-1][n-1]=True
        a,b,ans=0,0,0
        for l in range(n-2,-1,-1):
            for r in range(l,n):
                if l==r:
                    dp[l][r]=True
                elif l+1>r-1:
                    dp[l][r]=True if s[l]==s[r] else False
                else:
                    dp[l][r]=dp[l+1][r-1] and s[r]==s[l]
                if dp[l][r] and r-l>=ans:
                    a,b,ans=l,r,abs(l-r)
        return s[a:b+1]

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