[LeetCode] 0005. Longest Palindromic Substring

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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.

內容目錄

Python

				
					# time complexity: O(n)
# space complexity: O(n)
# Manacher's Algorithm
class Solution:
    def longestPalindrome(self, s: str) -> str:
        sPrime = "#" + "#".join(s) + "#"
        n = len(sPrime)
        palindromeRadii = [0] * n
        center = radius = 0
        for i in range(n):
            mirror = 2 * center - i
            if i < radius:
                palindromeRadii[i] = min(radius - i, palindromeRadii[mirror])
            while (
                i + 1 + palindromeRadii[i] < n
                and i - 1 - palindromeRadii[i] >= 0
                and sPrime[i + 1 + palindromeRadii[i]]
                == sPrime[i - 1 - palindromeRadii[i]]
            ):
                palindromeRadii[i] += 1
            if i + palindromeRadii[i] > radius:
                center = i
                radius = i + palindromeRadii[i]
        maxLength = max(palindromeRadii)
        centerIndex = palindromeRadii.index(maxLength)
        startIndex = (centerIndex - maxLength) // 2
        longestPalindrome = s[startIndex: startIndex + maxLength]

        return longestPalindrome


# Brute Force
# time complexity: O(n^3)
# space complexity: O(n^3)
class Solution:
    def longestPalindrome(self, s: str) -> str:
        def check(i, j):
            left = i
            right = j - 1
            while left < right:
                if s[left] != s[right]:
                    return False
                left += 1
                right -= 1
            return True
        for length in range(len(s), 0, -1):
            for start in range(len(s) - length + 1):
                if check(start, start + length):
                    return s[start: start + length]
        return ""

# time complexity: O(n^2)
# space complexity: O(n^2)
# Dynamic Programming
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]


s = "babad"
print(Solution().longestPalindrome(s))
s = "cbbd"
print(Solution().longestPalindrome(s))

Python

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