343. Integer Break(第十四周)
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Description:
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.
解题思路:
一开始觉得要尽可能多的乘数才会使最后的值更大,列出任意两个
n = 5 return 6 (5 = 3 + 2)
n = 8 return 18 (8 = 3 + 3 + 2)
因子要尽可能的多,我们发现因子要最好是3,因为4分成2+2得出的答案是一样的。即分解尽可能多的3。
有了以下:
class Solution {public: int integerBreak(int n) { if(n == 2) return 1; if(n == 3) return 2; int res = 1; while(n > 4) { res *= 3; n = n - 3; } return res * n; }};
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