2019年1月10日15:39:23 去掉了所有不必要的循环区间 还是超时 本地运行大概3s

#!/bin/python3

import math
import os
import random
import re
import sys # Complete the sherlockAndAnagrams function below.
def compareSubString(string1,string2):
num_string1 = [0]*26
num_string2 = [0]*26
for index in range (len(string1)):
temp = ord(string1[index]) - ord('a')
num_string1[temp] +=1
temp = ord(string2[index]) - ord('a')
num_string2[temp] +=1 if num_string1 == num_string2:
return True
else:
return False def sherlockAndAnagrams(s): #s <= 100
n = len(s)
count = 0
for index in range(n-1): #last one have no mean
for length in range(1, n - index): #only length between [1,n-1] passible have Anagrams
tempstring1 = s[index : index + length]
if (index + 1 <= n - length ):
for index2 in range(index + 1, n - length +1):
tempstring2 = s[index2: index2 + length]
if compareSubString(tempstring1 , tempstring2):
count +=1
return count if __name__ == '__main__':
# fptr = open(os.environ['OUTPUT_PATH'], 'w') q = 1 for q_itr in range(q):
s = 'zjekimenscyiamnwlpxytkndjsygifmqlqibxxqlauxamfviftquntvkwppxrzuncyenacfivtigvfsadtlytzymuwvpntngkyhw' result = sherlockAndAnagrams(s) print(result)

------

注意 anagrammatic  是拼字游戏的意思,任意调换顺序可以组成对方都成立,不是简单的回文数字。

------

1.第一种解法想省些时间 结果都比测试集少了20多个 应该是有些规则题目没有描述清楚。

5
ifailuhkqqhucpoltgtyovarjsnrbfpvmupwjjjfiwwhrlkpekxxnebfrwibylcvkfealgonjkzwlyfhhkefuvgndgdnbelgruel
gffryqktmwocejbxfidpjfgrrkpowoxwggxaknmltjcpazgtnakcfcogzatyskqjyorcftwxjrtgayvllutrjxpbzggjxbmxpnde
mqmtjwxaaaxklheghvqcyhaaegtlyntxmoluqlzvuzgkwhkkfpwarkckansgabfclzgnumdrojexnrdunivxqjzfbzsodycnsnmw
ofeqjnqnxwidhbuxxhfwargwkikjqwyghpsygjxyrarcoacwnhxyqlrviikfuiuotifznqmzpjrxycnqktkryutpqvbgbgthfges
zjekimenscyiamnwlpxytkndjsygifmqlqibxxqlauxamfviftquntvkwppxrzuncyenacfivtigvfsadtlytzymuwvpntngkyhw

399
471
370
403
428

#!/bin/python3

import math
import os
import random
import re
import sys # Complete the sherlockAndAnagrams function below.
def sherlockAndAnagrams(s):
n = len(s)
count = 0
# len = 1 && multi in string
for i in range(n):
for j in range(n):
if j > i:
if s[j] == s[i]:
count += 1
if j - i > 1:
count += 1 # repeat substring && huiwen substring
for start in range(n):
for end in range(n):
if end - start > 1: # substring len > 1
substring_num = 0
reverse_substring_num = 0
str1 = s[start:end]
str2 = s[end:]
str3 = str1[::-1]
substring_num = str2.count(str1)
reverse_substring_num = str2.count(str3)
count += substring_num
if str1 != str3: # avoid kkk
count += reverse_substring_num return count if __name__ == '__main__': s = 'zjekimenscyiamnwlpxytkndjsygifmqlqibxxqlauxamfviftquntvkwppxrzuncyenacfivtigvfsadtlytzymuwvpntngkyhw' result = sherlockAndAnagrams(s)
# fptr = open(os.environ['OUTPUT_PATH'], 'w')
#
# q = int(input())
#
# for q_itr in range(q):
# s = 'abba'
#
# result = sherlockAndAnagrams(s)
#
# fptr.write(str(result) + '\n')
#
# fptr.close()

2.看s 最大也就100 直接列出所有子串排序求和就行 但是发现会超时,本地运行需要十几秒。。。:

#!/bin/python3

import math
import os
import random
import re
import sys # Complete the sherlockAndAnagrams function below.
def sherlockAndAnagrams(s):
n = len(s)
count = 0
#find all substring
arr = []
for i in range(n):
for j in range(n+1):
if j > i:
arr.append(s[i:j])
# count
count = 0
for i in range(len(arr)):
for j in range(len(arr)):
if j>i and len(arr[i] )== len(arr[j]):
str1 = sorted(arr[i])
str2 = sorted(arr[j])
if str1 == str2 :
count += 1
return count if __name__ == '__main__': s = 'mqmtjwxaaaxklheghvqcyhaaegtlyntxmoluqlzvuzgkwhkkfpwarkckansgabfclzgnumdrojexnrdunivxqjzfbzsodycnsnmw' result = sherlockAndAnagrams(s)
# fptr = open(os.environ['OUTPUT_PATH'], 'w')
#
# q = int(input())
#
# for q_itr in range(q):
# s = 'abba'
#
# result = sherlockAndAnagrams(s)
#
# fptr.write(str(result) + '\n')
#
# fptr.close()

3.优化了下 改成字符串 还是超时:

#!/bin/python3

import math
import os
import random
import re
import sys # Complete the sherlockAndAnagrams function below.
def sherlockAndAnagrams(s):
n = len(s)
count = 0
#find all substring
arr = []
for i in range(n):
for j in range(n+1):
if j > i:
arr.append(s[i:j])
# count
count = 0
# for i in range(len(arr)):
# for j in range(len(arr)):
# if j>i and len(arr[i] )== len(arr[j]):
# str1 = sorted(arr[i])
# str2 = sorted(arr[j])
# if str1 == str2 :
# count += 1 for i in range(len(arr)):
temp = "".join(sorted(arr[i]))
arr[i] = temp for i in range(len(arr)):
for j in range(i,len(arr)):
if j>i and len(arr[i] )== len(arr[j]) and arr[i] == arr[j]:
count += 1
return count if __name__ == '__main__': s = 'ifailuhkqqhucpoltgtyovarjsnrbfpvmupwjjjfiwwhrlkpekxxnebfrwibylcvkfealgonjkzwlyfhhkefuvgndgdnbelgruel' result = sherlockAndAnagrams(s) print(result)
# fptr = open(os.environ['OUTPUT_PATH'], 'w')
#
# q = int(input())
#
# for q_itr in range(q):
# s = 'abba'
#
# result = sherlockAndAnagrams(s)
#
# fptr.write(str(result) + '\n')
#
# fptr.close()

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