LeetCode 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 and not larger than 58.
Hint:
There is a simple O(n) solution to this problem.
You may check the breaking results of n ranging from 7 to 10 to discover the regularities.
Amazing numeric question！！！看得hint试了一下才发现，把数字分解成n个3再加3，4，5的形式，能够让product最大。注意当n>=5时才满足这个规律。
public class Solution {
public int integerBreak(int n) {
if (n <= 2) return 1;
if (n == 3) return 2;
if (n == 4) return 4;
if (n == 5) return 6;
if (n == 6) return 9;
int div = n / 3;
int modular = n % 3;
if (modular == 0) {
return (int)Math.pow(3,div);
} else if (modular == 1) {
return (int)Math.pow(3,div - 1) * (3 + modular);
} else {
return (int)Math.pow(3,div) * modular;
}
}
}
