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Leetcode 343. Integer Break

2024-08-23 14:02:45   220人阅读

Given a positive integer n, break it into the sum of at least two positive integers and maximize the product of those integers. Return the maximum product you can get.

For example, given n = 2, return 1 (2 = 1 + 1); given n = 10, return 36 (10 = 3 + 3 + 4).

Note: You may assume that n is not less than 2 and not larger than 58.

 

一个大于3的数拆分的话，可以分成 n*3 + m (m=0,1,2) 这种形式来得到最大的乘积。这里面的例外是m=1的情况，因为这时，把这个1和最后一个3拆分成2+2，可以让product更大。

 1 class Solution(object):
 2     def integerBreak(self, n):
 3         """
 4         :type n: int
 5         :rtype: int
 6         """
 7         if n == 2:
 8             return 1
 9         if n == 3:
10             return 2
11             
12         prod = 1
13         
14         while n >= 3:
15             n -= 3
16             prod = 3*prod
17         
18         if n == 0:
19             return prod
20         elif n == 1:
21             return 4*prod/3
22         elif n == 2:
23             return 2*prod
24

 

Leetcode 343. Integer Break


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