算法训练营 -- 回文串
2019-06-08 本文已影响0人
拜仁的月饼
描述
给定一个字符串，求出该字符串有多少子串是回文串。
子串：字符串中连续的一段。比如字符串abcd里，bc、abc、a、bcd都是子串。
回文串：字符串倒序写出来和该字符串相同。比如aba，倒序写出来也是aba，故aba是回文串。而abab不是回文串，因为倒过来写是baba。
输入
输入一个字符串。
输出
输出子串是回文串的个数。
样例1输入
abab
样例1输出
6
样例1解释
a1，b1，a2
b2，aba，bab
提示
这篇文章是求最长的回文串的，那么如何求回文串的数目呢？可以发现manacher算法将每个位置为中心能延展出的最长回文串求出来了，那么这个最长回文串的一半（上取整）就是以该点作为中心的回文串数目。
注意答案要用long long。
Manacher算法详解
click here.
解法1：暴力求解(只能得50分)
#!/usr/bin/env pypy3
# encoding=utf-8
def palin(st): # This function is to judge whether the string "st" is a palindrome or not.
'''
:param st: the string that is judged
return : a boolean
'''
l = len(st) # The length of the string
a = True # Define a boolean variation whose defalt is True.
if l == 1: # If "st" consists of only one character:
return True # It is definitely a palindrome.
# ==============================================================
st_l = list(st) # Convert the string "st" to a list "st_l"
mid = l // 2 # The rank of the middle point
if l % 2 == 0: # If 'l' is a double number,
st_l.insert(mid, '#') # insert a sharp to the middle of the list "st_l"
# to make the length of the "st_l" a single number
l = len(st_l) # Update the value of "l"
# ==============================================================
i = mid - 1 # From middle to left
j = mid + 1 # From middle to right
while ((i >= 0) and (j < l)):
if st_l[i] != st_l[j]: # If asymmetirc happened,
a = False # "a" becomes "False".
break # Break the loop.
else:
i -= 1
j += 1
return a
def getAnswer(Str):
le = len(Str)
cnt = 0 # Count
for i in range(le): # Start
for j in range(i + 1, le + 1):
h = palin(Str[i: j])
if (h == True):
cnt += 1
return cnt
if __name__ == "__main__":
print(getAnswer(input()))
