leetcode5 Longest Palindromic Substring

"""
Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.
Example 1:
Input: "babad"
Output: "bab"
Note: "aba" is also a valid answer.
Example 2:
Input: "cbbd"
Output: "bb"
"""

"""
解法一：中心向两边扩张法
分了奇数和偶数两种情况
当s[l] == s[r]
l--, r++
"""
class Solution:
    def longestPalindrome(self, s):
        res = ''
        for i in range(len(s)):
            # odd  'aba'
            temp = self.palindromic(s, i, i)
            if len(res) < len(temp):
                res = temp
            # even 'abba'
            temp = self.palindromic(s, i, i+1)
            if len(res) < len(temp):
                res = temp
        return res
    # get the longest palindrome, l, r are the middle indexes
    # from inner to outer
    def palindromic(self, s, l, r):
        while l >= 0 and r <= len(s) - 1 and s[l] == s[r]:
            l -= 1
            r += 1
        return s[l+1: r]
"""
用动态规划做
"""
class Solution(object):
    def longestPalindrome(self, s):
        if s is None:
            return ''
        ret = ''
        lens = len(s)
        max = 0
        dp = [[0] * lens for i in range(lens)]
        #产生一个lens*lens全0二维数组，dp数组后面存储True或False
        for j in range(lens):
            for i in range(j + 1):
                dp[i][j] = ((s[i] == s[j]) and (j - i <= 2 or dp[i + 1][j - 1]))
                #如果s[i]=s[j]说明串的首尾相同，
                # 并且j-i为0表示只有一个字符必为回文，
                # j-i=1两个字符切相等必为回文，
                # j-i=2三个字符首尾相同无论中间是什么必为回文，
                # 或者dp[i + 1][j - 1]为真表示去掉首尾为回文，而新加的首尾相同必为回文。
                if dp[i][j] and j - i + 1 > max:
                    max = j - i + 1
                    ret = s[i:j + 1]
                    #表示i开头j结束的串回文并且最长则更新长度max和回文串ret。
        return ret