LeetCode005. 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"

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思路：以每一个字母为中心，向两边寻找；或者以两个字母为中心向两边寻找。

class Solution(object):
    def longestPalindrome(self, s):
        """
        :type s: str
        :rtype: str
        """
        maxi=0
        l=0
        r=0
        def helper(s,l,r):
            while l>=0 and r<len(s) and s[l]==s[r]:
                l -= 1
                r += 1
            length = r-l+1
            return l+1,r-1,length
        
        for i in range(len(s)):
            l1, r1, length1 = helper(s,i,i)
            l2, r2, length2 = helper(s,i,i+1)
            if length1>maxi:
                maxi = length1
                l,r=l1,r1
            if length2>maxi:
                maxi = length2
                l,r=l2,r2
        return s[l:r+1]

第一次写的时候两边界线出错。 在helper写中的是while l>0.这样从第一个字母开始回文就会出错。另外l>=0时，对应最后的s[l+1,r]

还有用dp算法,比较复杂,详情参考https://blog.youkuaiyun.com/L141210113/article/details/87917988

class Solution(object):
    def longestPalindrome(self, s):
        if len(s)<2: return s
        maxlen=0
        result=''
        dp = [[0]*len(s) for i in range(len(s))]
        for j in range(len(s)):
            for i in range(j+1):
                dp[i][j] = (s[i]==s[j]) and (j-i<=2 or dp[i+1][j-1])
                if dp[i][j]:
                    if maxlen < j-i+1:
                        maxlen = j-i+1
                        result = s[i:j+1]
        return result