最近再刷leetcode,除了链表之外的都用python 实现,贴出一些代码,希望指正.
问题描述:
Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.
字符串中最长的回文数
Input: "babad"
Output: "bab"
Note: "aba" is also a valid answer.
Input: "cbbd"
Output: "bb"
解决方案.
遍历字符串,拿到一个字符,从后往前找相同的字符的位置,返回所有相同的位置索引列表,计算两个相同的字符能不能构成回文数.
用babad举例.
从b开始遍历,从abad中从后往前找b,当然这里只有一个b所以只会返回一个b的下标是2,然后计算下标从0到2是不是构成回文数,这里刚好是,计算长度,len = 3.下一个
然后遍历a同理,找到相同字符下标的位置3,那么判断下标1和3是不是构成回文数,这里也是 len = 3,继续向下遍历.
用能aaabaaaa举例.
从a开始遍历,从a之后的aabaaaa中从后往前找和a相同的字符的位置,这里相同的比较多,返回list = [7,6,5,4,2,1],那么先比较下标0到7是不是构成回文数,aaabaaaa后面的a比前面多一个,所以不是回文数,然后看0到6是不是回文数,aaabaaa刚好是,则终止判断,len = 7 继续向后遍历.
注意: python 中判断是不是回文数 return list == list[::-1]即可,list[::-1]表示翻转list
class Solution(object):
def longestPalindrome(self, s):
"""
:type s: str
:rtype: str
"""
def is_palindromic(list):
return list == list[::-1], len(list), list
def find_next(list1, tar):
# print(list1)
list3 = []
list2 = list1[::-1]
# print(list2)
for i, each in enumerate(list2):
if each == tar:
list3.append(len(list1) - i)
return list3
max = 0
SUBlist = ""
if len(s) == 1:
return s
for i, each in enumerate(s):
same_char_list = find_next(s[i + 1:], each)
if same_char_list is not None and len(same_char_list) != 0:
for each in same_char_list:
# print(i, i + each)
flag, lenght, sublist = is_palindromic(s[i:1 + i + each])
if flag:
if max < lenght:
max = lenght
SUBlist = sublist
if lenght == len(s):
return SUBlist
else:
if SUBlist == "":
SUBlist = each
return SUBlist
solution = Solution()
print(solution.longestPalindrome("aaabaaaa"))
