343. Integer Break

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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.

给一个大于2的整数，将这个整数分成两个以上数字的和，使他们乘积最大

python answer 1:

class Solution(object):    def integerBreak(self, n):        """        :type n: int        :rtype: int        """        if n <= 3:            return n-1        if (n-2) % 3 == 0:            return 3**((n-2)/3)*2        if (n-3) % 3 == 0:            return 3**(n/3)        if (n-4) % 3 == 0:            return 3**((n-4)/3)*4

Python answer 2:

class Solution(object):    def integerBreak(self, n):        """        :type n: int        :rtype: int        """        dp = [0, 0, 1]        for i in range(3, n+1):            dp.append(0)            for j in range(i):                dp[i] = max(dp[i], max(dp[i-j], i-j)*max(dp[j], j))             return dp[n]

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