將s翻轉成s1 求s和s1的最長公共子序列長度
可以先判斷下s是否是迴文
另外一個思路
dp[i][j] = dp[i + 1][j - 1] + 2 if s[i] == s[j]
dp[i][j] = max(dp[i][j - 1], dp[i + 1][j]) otherwise

class Solution(object):
    def longestPalindromeSubseq(self, s):
        """
        :type s: str
        :rtype: int
        """
 

        size = len(s)
        s1 = s[::-1]
        if s1 == s: return size
        dp = [[0]*(size+1) for i in range(size+1)]
        for i in range(size):
            for j in range(size):
                if s[i]==s1[j]: 
                    dp[i+1][j+1] = dp[i][j]+1
                else:
                    dp[i+1
 
][j+1] = max(dp[i][j+1],dp[i+1][j])
        return dp[size][size]

在每個字元的兩邊都插入一個特殊的符號’#’
定義 id 為已知的 {右邊界最大} 的迴文子串的中心，mx則為id+P[id]，也就是這個子串的右邊界

class Solution(object):
    def longestPalindrome(self,s):
        '''
        :param s: str
        :return: str
        '''
        ss = '$'
 
;idx=0;mx=0
        for i in s:
            ss = ss+'#'+i
        ss = ss+'#'
        P = [0 for i in range(len(ss))]
        for i in range(1,len(ss)):
            if P[idx]+idx<i:
                P[i] = 1
            else:
                P[i] = min(P[2*idx-i],P[idx]+idx-i) #對稱點
            while(i+P[i]<len(ss) and ss[i-P[i]]==ss[i+P[i]]):
                P[i] = P[i]+1
            if idx + P[idx] < i + P[i]:
                idx = i
            if P[i]>P[mx]:
                mx = i
        a = ss[mx - (P[mx] - 1):mx + P[mx]]
        ret = ''
        for i in a:
            if i!='#':
                ret+=i
        return ret

判斷s[i:j+1]是否是迴文等價於判斷s[i+1:j] 和 s[i][j]的情況

class Solution(object):
    def countSubstrings(self, s):
        """
        :type s: str
        :rtype: int
        """
        if s=='': return 0
        size = len(s)
        dp = [[0 for i in range(size)] for j in range(size)]
        ret = 0
        for i in range(size):
            dp[i][i]=1
            for j in range(i):
                if s[i]==s[j]:
                    if dp[i-1][j+1] or j==i-1:
                        dp[i][j] = 1
            ret += sum(dp[i])
        return ret