【LeetCode】整数拆分

问题描述

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.

• Example 1:
  Input: 2
  Output: 1
  Explanation: 2 = 1 + 1, 1 × 1 = 1.

• Example 2:
  Input: 10
  Output: 36
  Explanation: 10 = 3 + 3 + 4, 3 × 3 × 4 = 36.

Note: You may assume that n is not less than 2 and not larger than 58.
转至具体题目

解题思路

题目要求输入一个正整数n，将其拆分为至少2个正整数之和，使其乘积最大。很容易想到，数字较小时，将其分为相等的2个数乘积最大。参考网上的数学证明后发现，拆出的因子如果不限于整数，则等于e=2.718…是最佳选择（具体证明过程可以百度），因为该题限定正整数，所以我们只要拆出尽可能多的3即可

代码块

class Solution {
public:
    int integerBreak(int n) {
        if (n == 2 || n == 3) return n - 1;
        if (n == 4) return 4;
        n -= 5;
        return (int)pow(3, (n / 3 + 1)) * (n % 3 + 2);
    }
};