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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.
Hint:
规律:大于4时,不断寻找因子3,并将其相乘。
// Runtime: 0 ms
public class Solution {
public int integerBreak(int n) {
if (n == 2 || n == 3) {
return n - 1;
}
int base = 1;
while (n > 4) {
base *= 3;
n -= 3;
}
return base * n;
}
}
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原文地址:http://blog.csdn.net/foreverling/article/details/51419523