Python Leetcode之回文串相关

最新推荐文章于 2024-11-10 15:41:12 发布

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最新推荐文章于 2024-11-10 15:41:12 发布

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文章标签： python leetcode

本文链接：https://blog.youkuaiyun.com/qq_36103746/article/details/115802256

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本文详细解析了回文子串与回文子序列的算法实现，包括动态规划与中心扩展法两种方法，并提供了Python代码示例。通过本文，读者可以深入理解回文相关算法的设计思想。

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回文子串是要连续的，回文子序列可以不是连续的！回文子串、回文子序列都是动态规划经典题目。

回文子串

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题目讲解

class Solution:
    def countSubstrings(self, s: str) -> int:
        if not s: return 0
        n = len(s)
        dp = [[False] * n for _ in range(n)]
        res = 0
        for i in range(n - 1, -1, -1):
            for j in range(i, n):
                if s[i] == s[j] and (j - i < 2 or dp[i + 1][j - 1]):
                    res += 1
                    dp[i][j] = True   
        return res

最长回文子串

在这里插入图片描述
该题与上一题类似，只是记录最长的那一个回文子串，代码整体一致。

class Solution:
    def longestPalindrome(self, s: str) -> str:
        n = len(s)
        dp = [[False] * n for _ in range(n)]
        out_s = temp = str()
        for i in range(n - 1, -1, -1):
            for j in range(i, n):
                if s[i] == s[j] and (j - i <= 1 or dp[i + 1][j - 1]):
                    dp[i][j] = True
                    temp = s[i:j + 1]
                    if len(temp) >= len(out_s): out_s = temp
                else:
                    temp = str()
        return out_s

以上两题也可以采用中心扩展法，具体讲解参照两道回文子串的解法（详解中心扩展法）

# 回文子串
class Solution:
    def countSubstrings(self, s: str) -> int:
        if not s: return 0
        n = len(s)
        res = 0
        for i in range(2 * n - 1):
            l = i // 2
            r = l + i % 2
            while l >= 0 and r < n and s[l] == s[r]:
                res += 1
                l -= 1
                r += 1
        return res

# 最长回文子串
class Solution:
    def longestPalindrome(self, s: str) -> str:
        n = len(s)
        out_s = temp = str()
        for i in range(2 * n - 1):
            l = i // 2
            r = l + i % 2
            while l >= 0 and r < n and s[l] == s[r]:
                temp = s[l : r + 1]
                l -= 1
                r += 1
                if len(temp) >= len(out_s): out_s = temp
        return out_s

# 也可以写成以下形式：
class Solution:
    def countSubstrings(self, s: str) -> int:
        n = len(s)
        res = 0
        for i in range(n):
            res += self.funer(s, i, i, 0)
            res += self.funer(s, i, i + 1, 0)
        return res
    def funer(self, s, i, j, res):
        while i >= 0 and j < len(s) and s[i] == s[j]:
            i -= 1;j += 1;res += 1
        return res

class Solution:
	def longestPalindrome(self, s):
	    res = ""
	    for i in range(len(s)):
	        # odd case, like "aba"
	        tmp = self.helper(s, i, i)
	        if len(tmp) > len(res):
	            res = tmp
	        # even case, like "abba"
	        tmp = self.helper(s, i, i+1)
	        if len(tmp) > len(res):
	            res = tmp
	    return res	 
	# get the longest palindrome, l, r are the middle indexes   
	# from inner to outer
	def helper(self, s, l, r):
	    while l >= 0 and r < len(s) and s[l] == s[r]:
	        l -= 1; r += 1
	    return s[l+1:r]

最长回文子序列

在这里插入图片描述

题目讲解

class Solution:
    def longestPalindromeSubseq(self, s: str) -> int:
        n = len(s)
        dp = [[0] * n for _ in range(n)]
        for i in range(n):
            dp[i][i] = 1
        for i in range(n - 1, -1, -1):
            for j in range(i + 1, n):
                if s[i] == s[j]:
                    dp[i][j] = dp[i + 1][j - 1] + 2
                else:
                    dp[i][j] = max(dp[i + 1][j], dp[i][j - 1])
        return dp[0][-1]