题解 | #最长回文子串#

#用dp数组存储了 较小长度回文串的数据 较大的直接在较短的上面继续增加 所以少了一层循环
s = input()
maxlen = 0
ls = len(s)
dp = [[0]*len(s) for _ in range(len(s))]
# 长度为1的回文串
for i in range(ls):
    dp[i][i] = 1

maxlen = 1
# 长度为2的回文串
for i in range(ls-1):
    if s[i] == s[i+1]:
        dp[i][i+1] = 1
        maxlen = 2
# 长度为l的回文串

for L in range(3,ls+1):
    find_flag = False
    for i in range(ls-L+1):
        j = i+L-1
        if s[i] == s[j]:
            dp[i][j] = dp[i+1][j-1]
            # 存在长为L回文串
            if dp[i][j] == 1:
                find_flag = True
                maxlen =L

print(maxlen)