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 question２泊妗Ｗ饽弧！看得hint試了一下才發(fā)現(xiàn)谈息，把數(shù)字分解成n個3再加3域蜗，4巨双，5的形式，能夠讓product最大霉祸。注意當(dāng)n>=5時才滿足這個規(guī)律炉峰。
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;
}
}
}
