题解 | #密码截取#

原理

字符串的操作判断，求最长回文串。
回文串最基本就两种：ABA型、ABBA型（奇数串和偶数串），两边循环遍历判断长度即可。时间复杂度O(n^2)

my code

while True:
    try:
        ss = input()
        cnt = []
        for i in range(len(ss)):
            if i == len(ss)-1:
                break
            if ss[i] == ss[i+1]:
                n, x = 2, 1
                while i-x>=0 and i+x+1<len(ss):
                    if ss[i-x] == ss[i+x+1]:
                        n+=2
                        x+=1
                    else:
                        break
                cnt.append(n)
            elif i!=0 and ss[i-1] == ss[i+1]:
                n, y = 1, 1
                while i-y>=0 and i+y<len(ss):
                    if ss[i-y] == ss[i+y]:
                        n+=2
                        y+=1
                    else:
                        break
                cnt.append(n)
        print(max(cnt))
    except:
        break