[Leetcode] [Python] Longest palindromic Substring

#-*-Coding:utf8-*-
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__author__ = ' [email protected] '
https://oj.leetcode.com/problems/longest-palindromic-substring/

Given A string S, find the longest palindromic substring in S. Assume that the maximum length of S is 1000,
And there exists one unique longest palindromic substring.

===comments by dabay===
Start judging from the longest string until the length is 2. However, such an algorithm will timeout time Limit exceeded.
Using a well-known algorithm called Manacher ' s algorithm:
Each character is separated by a # to be traversed, and the length of the oldest string is recorded from each character to the opposite side.
Add # Here to avoid the problem that palindrome strings are odd or even.
Finally, in the resulting array of record lengths, you can find the maximum palindrome string.

S # a # b # b # # a # b # C # b # a #
P 1 2 1 2 5 2 1 4 1 2 1 6 1 2 1 2 1

The mid record has the center position of the palindrome substring with the largest right boundary, and Max_right records the right edge of the maximum substring.

Initialize position 0

For each position I
If I do not exceed max_right:
Judging if the palindrome length based on the mid symmetry point exceeds the boundary
If the boundary is not exceeded:
Do not continue to judge, the direct assignment is the value of this symmetry point, continue to judge the next I
If the boundary is exceeded:
The part within the boundary does not have to be judged (because of symmetry, it must be a palindrome), continue to look for the longest palindrome, record length
If I exceeds the max_right:
Look for the longest palindrome, record length

If Max_right has been expanded, update mid and Max_right
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Class Solution:
# @return A string
def longestpalindrome (self, s):
S_list = [' # ']
For C in S:
S_list.append (c)
S_list.append (' # ')

Max_right = 0
MID = 0
p = [0 for _ in S_list]
P[0] = 1

For I in xrange (1, Len (P)):
If I <= max_right:
If P[2*mid-i] < max_right-i:
P[i] = P[2*mid-i]
Continue
Else
j = Max_right-i
While I + J < Len (s_list) and i-j >= 0 and s_list[i+j] = = S_list[i-j]:
j = j + 1
P[i] = j
Else
j = 1
While I + J < Len (s_list) and i-j >= 0 and s_list[i+j] = = S_list[i-j]:
j = j + 1
P[i] = j
If Max_right < P[i] + i-1:
Max_right = P[i] + i-1
MID = I

max_p = max (P)
Max_index = P.index (max_p)

Longest = ""
For I in range (max_index-max_p+1, max_index+max_p):
If s_list[i] = = ' # ':
Continue
Longest = longest + s_list[i]
return longest


def main ():
s = solution ()
String = " CUBWQVHXDAMMPKWKYCRQTEGEPYXTOHSPEASRDTINJHBESILSVFFMNZZNMLTSSPJWUOGDYZVANALOHMZRYWDWQQCUKJCEOTHYDLGTOCUKC "
#string = "ABBABCBA"
Print S.longestpalindrome (String)


if __name__ = = "__main__":
Import time
Start = Time.clock ()
Main ()
Print "%s sec"% (Time.clock ()-start)

[Leetcode] [Python] Longest palindromic Substring