[leetcode]5. 最长回文子串

最新推荐文章于 2025-08-11 22:22:04 发布

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最新推荐文章于 2025-08-11 22:22:04 发布

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文章标签： python 数据结构与算法

原文链接：https://my.oschina.net/bugyang/blog/1935825

博客围绕Python展开，提出给定字符串s，求其最长回文子串的问题，假设s最大长度为1000。还介绍了多种解法，包括暴力破解、动态规划、O(n)解法以及利用字符串翻转比对获取最长回文等。

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2019独角兽企业重金招聘Python工程师标准>>>

给定一个字符串 s，找到 s 中最长的回文子串。你可以假设 s 的最大长度为1000。

示例 1：

输入: "babad"
输出: "bab"
注意: "aba"也是一个有效答案。

示例 2：

输入: "cbbd"
输出: "bb"

解法一：暴力破解，时间复杂度n^3

class Solution(object):
    def longestPalindrome(self, s):
        """
        :type s: str
        :rtype: str
        """
        s_len = len(s)
        max_str = ""
        max_len = 0
        tmp_str = ""
        tmp_len = 0
        round_str_dict = {}
        for i in range(0, s_len - 1):
            tmp_str = s[i]
            tmp_len = 1
            for j in range(i + 1, s_len):
                tmp_str += s[j]
                tmp_len += 1
                if not round_str_dict.has_key(tmp_str):
                    round_str_dict[tmp_str] = self.is_round_str(tmp_str)
                if round_str_dict[tmp_str]:
                    if tmp_len > max_len:
                        max_len = tmp_len
                        max_str = tmp_str
        if s_len !=0 and max_str == "":
            max_str = s[0]
        return max_str
                
            
    def is_round_str(self, tmp_str):
        tmp_str_len = len(tmp_str)
        if tmp_str_len == 0:
            return True
        i = 0
        j = tmp_str_len - 1
        is_round_str = True
        while i < j:
            if tmp_str[i] == tmp_str[j]:
                i += 1
                j -= 1
            else:
                is_round_str = False
                break
        return is_round_str

方法二：动态规划

class Solution(object):
    def longestPalindrome(self, s):
        """
        :type s: str
        :rtype: str
        """
        s_len = len(s)
        max_len = 0
        max_str = ""
        round_list = [[0 for i in range(0, s_len)] for j in range(0, s_len)]
        for j in range(0, s_len):
            for i in range(0, j + 1):
                if j - i <= 1:
                    if (s[i] == s[j]):
                        round_list[i][j] = 1
                        if (j - i + 1) > max_len:
                            max_len = j - i + 1
                            max_str = s[i:j + 1]
                    else:
                        round_list[i][j] = 0
                else:
                    if (s[i] == s[j]) and round_list[i + 1][j - 1] == 1:
                        round_list[i][j] = 1
                        if (j - i + 1) > max_len:
                            max_len = j - i + 1
                            max_str = s[i:j + 1]
                    else:
                        round_list[i][j] = 0
        return max_str

三、O(n)解法，参考https://segmentfault.com/a/1190000003914228

class Solution(object):
    def longestPalindrome(self, s):
        """
        :type s: str
        :rtype: str
        """
        m_id = 0
        max_id = 0
        max_p_id = 0
        s_tmp = "#" + "#".join(s) + "#"
        s_tmp_len = len(s_tmp)
        p_list = [0 for i in range(0, s_tmp_len)]
        print p_list
        for i in range(0, s_tmp_len):
            # print "%s %s %s %s" % (s_tmp[i], i, m_id, max_id)
            p_list[i] = min(p_list[2*m_id - i], max_id - i) if max_id > i else 1
            while (i - p_list[i] >= 0 and i + p_list[i] < s_tmp_len) \
            and s_tmp[i + p_list[i]] == s_tmp[i - p_list[i]]:
                p_list[i] += 1
            if p_list[i] > max_p_id:
                m_id = i
                max_id = i + p_list[i] - 1
                max_p_id = p_list[i]
        # return "".join(s[m_id - max_id : m_id + max_id].split("#"))
        # print p_list
        # print "%s %s" % (m_id, max_p_id) 
        return "".join(s_tmp[m_id-max_p_id+1:max_id].split("#"))

四、利用字符串翻转比对，获取最长回文

class Solution(object):
    def longestPalindrome(self, s):
        """
        :type s: str
        :rtype: str
        """
        n = len(s)
        maxl = 0
        start = 0
        for i in xrange(n):
            if i - maxl >= 0 and s[i-maxl: i+1] == s[i-maxl: i+1][::-1]:
                start = i - maxl
                maxl += 1
                print "a  %s %s %s" % (i, start, maxl)
                continue
            if i - maxl >= 1 and s[i-maxl-1: i+1] == s[i-maxl-1: i+1][::-1]:
                start = i - maxl - 1
                maxl += 2
                print "b  %s %s %s" % (i, start, maxl)
        return s[start: start + maxl]

转载于:https://my.oschina.net/bugyang/blog/1935825