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"

解题思路1 比较简单易懂，不过对于连续重复的字符串存在耗时操作

解题思路2 是将连续重复字符串看成一个整体了，比如 abbba 下表是01234   相同的不比较了，直接比较相同的左边跟右边

class Solution:
    def getTmpAns(self,s,l,r):
        size = len(s)
        while l>=0 and r<size and s[l] == s[r]:
            l = l-1
            r = r+1
        return s[l+1:r]
    def longestPalindrome(self, s):
        """
        :type s: str
        :rtype: str
        """
        ans = ""
        for i in range(len(s)):
            tmpAns = self.getTmpAns(s,i,i)
            if len(tmpAns) > len(ans):
                ans = tmpAns
            tmpAns = self.getTmpAns(s,i,i+1)
            if len(tmpAns) > len(ans):
                ans = tmpAns
        return ans
    
        def _longestPalindrome(self, s):
        """
        :type s: str
        :rtype: str
        """
        size = len(s)
        if size == 1 or size == 0:
            return s
        if size == 2:
            if s[0] == s[1]:
                return s
            else:
                return s[0]
        i = 0
        maxp = 1
        ans = s[0]
        while i < size:
            j = i+1
            while j < size: # let same num as a item  like bbb，i是左边起始位置 j是右边起始位置
                if s[i] == s[j]:
                    j += 1
                else:
                    break
            k = 0
            while i-k-1>=0 and j+k<size:
                if s[i-k-1] != s[j+k]:
                    break
                k += 1
            if j+k-i+k > maxp:
                maxp = j+k-i+k
                ans = s[i-k:j+k]
            if j+k == size-1:
                break
            i = j
        return ans