给出两个数值X和Y，返回在这个区间里的回文数，并且要求它们的平方根也是回文数。
例如（X=1，Y=10000）：
[(1, 1), (2, 4), (3, 9), (11, 121), (22, 484), (101, 10201), (111, 12321), (121, 14641), (202, 40804), (212, 44944)]
代码一：
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
def isPalindrome(n):
s = str(n)
return s == s[::-1]
def main():
m1 = 1
m2 = 10000000
return [(i, i*i) for i in range(m1,m2) if isPalindrome(i) and isPalindrome(i*i)]
if __name__ == '__main__':
import time
st = time.time()
print(main(), time.time() - st, sep='\n')
代码二：算法有进步，速度没改进
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
def isPalindrome(n):
s = str(n)
return s == s[::-1]
def filterNum(n):
s = str(n)
if int(s[-1]) not in {1,2,3}:
return False
else:
return s == s[::-1]
def main(n):
return [(i, i*i) for i in range(1,n) if filterNum(i) and isPalindrome(i*i)]
if __name__ == '__main__':
import time
st = time.time()
print(main(10000000), time.time()-st, sep='\n')
代码三：这个速度最快
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
from time import time
def filterNum(n):
s = str(n)
return s == s[::-1] if s[-1] in '123' else False
def main(n):
isPalindrome = lambda n: str(n) == str(n)[::-1]
return [(i, i*i) for i in range(1,n) if filterNum(i) and isPalindrome(i*i)]
if __name__ == '__main__':
st = time()
print(main(10000000), time()-st, sep='\n')
