第五题 回文串

第一思路 先写个判断回文的函数

def judge(list1):
    print(list1[0:0])
    low = 0
    high = len(list1)-1
    if high <= 0:
        return True
    while low < high:
        if list1[low] == list1[high]:
            low += 1
            high -= 1
        else:
            break
    if low <high or list1[low] != list1[high]:
        return False
    else:
        return True

然后以每个字母为头去判断回文长度 

发现根本不靠谱 因为有的短的不是回文 长的却是  这时间复杂度上天了

下面这个是错的 

def long(s):
    length = 1
    maxlen = 1
    for i,each in enumerate(s):
        while i+length < len(s):
            if judge(s[i:i+length]) :
                length += 1
            else:
                break
        if length-1 > maxlen:
            maxlen = length-1
        length = 1

    return maxlen

print(long('aba'))

 

后面想着以每个回文中心作为判断依据

 

 

但是要注意 中心可能在两个字母中间  所以有了以下的代码

:

def judge2(s):
    i = 0
    high = len(s)-1
    maxlen = 0
    maxpos = 0
    while i <= high:
        if i % 1 == 0:
            i = int(i)
            j = 0
            while (i-j)>=0 and (i+j)<=high:
                if s[i-j] == s[i+j]:
                    length = 1+j*2
                else:
                    break
                if length > maxlen:
                    maxpos = int(i)
                    maxlen = length
                j += 1

        else:
            j = 0
            ilow = int(i//1)
            while (ilow-j)>=0 and (ilow+1+j)<=high:
                if s[ilow-j] == s[ilow+1+j]:
                    length = j*2+2
                else:
                    break
                if length > maxlen:
                    maxpos = int(ilow)
                    maxlen = length
                j += 1

        i = i +0.5
    maxpos = int(maxpos)
    maxlen = int(maxlen)
    if maxlen %2 == 0:
        return s[maxpos-maxlen//2+1:maxpos+maxlen//2+1]
    else:
        return s[maxpos-maxlen//2:maxpos+maxlen//2+1]

 

每次步进0.5往后便历  

如果不是整数  说明这一步在判断偶数个字母组成的回文.

 

标签：int,length,ilow,maxlen,第五,maxpos,回文
来源： https://www.cnblogs.com/xiaoli1996/p/15076137.html