python中3位数的最高回文数
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highest palindrome with 3 digit numbers in python
提问by FabianCook
In problem 4 from http://projecteuler.net/it says:
在http://projecteuler.net/ 的问题 4 中，它说：
A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 * 99.
Find the largest palindrome made from the product of two 3-digit numbers.
回文数的读法是一样的。由两个两位数的乘积构成的最大回文数是 9009 = 91 * 99。
找出由两个 3 位数字的乘积构成的最大回文数。
I have this code here
我这里有这个代码
def isPalindrome(num):
return str(num) == str(num)[::-1]
def largest(bot, top):
for x in range(top, bot, -1):
for y in range(top,bot, -1):
if isPalindrome(x*y):
return x*y
print largest(100,999)
It should find the largest palindrome, it spits out 580085which I believe to be correct, but project euler doesn't think so, do I have something wrong here?
它应该找到最大的回文，它吐出580085我认为是正确的，但项目 euler 不这么认为，我这里有什么问题吗？
When I revered the for loop I didn't think it through, I removed the thing that checks for the biggest, silly me. Heres the working code
当我敬畏 for 循环时，我没有考虑清楚，我删除了检查最大、最愚蠢的我的东西。这是工作代码
def isPalindrome(num):
return str(num) == str(num)[::-1]
def largest(bot, top):
z = 0
for x in range(top, bot, -1):
for y in range(top,bot, -1):
if isPalindrome(x*y):
if x*y > z:
z = x*y
return z
print largest(100,999)
it spits out 906609
它吐出 906609
采纳答案by John Kugelman
Iterating in reverse doesn't find the largest x*y, it finds the palindrome with the largest x. There's a larger answer than 580085; it has a smaller xbut a larger y.
反向迭代不会找到最大的x*y，它会找到最大的回文x。有一个比 580085 更大的答案；它有一个较小x但较大的y.
回答by Jon Clements
This would more efficiently be written as:
这将更有效地写为：
from itertools import product
def is_palindrome(num):
return str(num) == str(num)[::-1]
multiples = ( (a, b) for a, b in product(xrange(100,999), repeat=2) if is_palindrome(a*b) )
print max(multiples, key=lambda (a,b): a*b)
# (913, 993)
You'll find itertoolsand generators very useful if you're doing Euler in Python.
itertools如果您在 Python 中执行 Euler，您会发现生成器非常有用。
回答by blueberryfields
Tried making it more efficient, while keeping it legible:
尝试使其更高效，同时保持清晰：
def is_palindrome(num):
return str(num) == str(num)[::-1]
def fn(n):
max_palindrome = 1
for x in range(n,1,-1):
for y in range(n,x-1,-1):
if is_palindrome(x*y) and x*y > max_palindrome:
max_palindrome = x*y
elif x * y < max_palindrome:
break
return max_palindrome
print fn(999)
回答by Rudney
ReThink: efficiency and performance
重新思考：效率和性能
def palindrome(n):
maxNumberWithNDigits = int('9' * n) #find the max number with n digits
product = maxNumberWithNDigits * maxNumberWithNDigits
#Since we are looking the max, stop on the first match
while True:
if str(product) == str(product)[::-1]: break;
product-=1
return product
start=time.time()
palindrome(3)
end=time.time()-start
palindrome...: 997799, 0.000138998031616 secs
回文...：997799, 0.000138998031616 秒
回答by user3833942
Not the most efficient answer but I do like that it's compact enough to fit on one line.
不是最有效的答案，但我确实喜欢它足够紧凑，可以放在一条线上。
print max(i*j for i in xrange(1,1000) for j in xrange(1,1000) if str(i*j) == str(i*j)[::-1])
回答by CooLee
Here I added two 'break' to improve the speed of this program.
这里我添加了两个'break'来提高这个程序的速度。
def is_palindrome(num):
return str(num) == str(num)[::-1]
def max_palindrome(n):
max_palindrome = 1
for i in range(10**n-1,10**(n-1)-1,-1):
for j in range(10**n-1,i-1,-1):
if is_palindrome(i*j) and i*j > max_palindrome:
max_palindrome = i * j
break
elif i*j < max_palindrome:
break
return max_palindrome
n=int(raw_input())
print max_palindrome(n)
回答by kaffuffle
Simple:
简单的：
def is_pallindrome(n):
s = str(n)
for n in xrange(1, len(s)/2 + 1):
if s[n-1] != s[-n]:
return False
return True
largest = 0
for j in xrange(100, 1000):
for k in xrange(j, 1000):
if is_pallindrome(j*k):
if (j*k) > largest: largest = j*k
print largest
回答by Anoop M
Each time it doesnot have to start from 999 as it is already found earlier.Below is a simple method using string function to find largest palindrome using three digit number
每次都不必从 999 开始，因为它之前已经找到了。 下面是一个使用字符串函数使用三位数字查找最大回文的简单方法
def palindrome(y):
z=str(y)
w=z[::-1]
if (w==z):
return 0
elif (w!=z):
return 1
h=[]
a=999
for i in range (999,0,-1):
for j in range (a,0,-1):
l=palindrome(i*j)
if (l==0):
h=h+[i*j]
a-=1
print h
max=h[0]
for i in range(0,len(h)):
if (h[i] > max):
max= h[i]
print "largest palindrome using multiple of three digit number=%d"%max
回答by Darshi
Here is my code to solve this problem.
这是我解决这个问题的代码。
lst = []
for i in range(100,1000):
for n in range(2,i) :
lst.append (i* n)
lst.append(i*i)
lst2=[]
for i in lst:
if str(i) == str(i)[::-1]:
lst2.append(i)
print max(lst2)
回答by CSEKU_150236
580085 = 995 X 583, where 906609 = 993 X 913 found it only by applying brute forcing from top to bottom!
580085 = 995 X 583，其中 906609 = 993 X 913 只能通过从上到下应用蛮力找到它！
