LeetCode题目：343. Integer Break-优快云博客

本文链接：https://blog.youkuaiyun.com/sinat_37607788/article/details/72586174

探讨了如何将一个正整数拆分成至少两个正整数之和，并最大化这些整数的乘积。通过数学分析得出最优解的特性，即尽可能多地使用3进行拆分，并给出O(N)的时间复杂度解决方案。

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题目描述：

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.

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.

参考大神的解释，一下子明白了：

When x=N/2, we get the maximum of this function.

However, factors should be integers. Thus the maximum is (N/2)*(N/2) when N is even or (N-1)/2 *(N+1)/2 when N is odd.

When the maximum of f is larger than N, we should do the break.

(N/2)*(N/2)>=N, then N>=4

(N-1)/2 *(N+1)/2>=N, then N>=5

These two expressions mean that factors should be less than 4, otherwise we can do the break and get a better product. The factors in last result should be 1, 2 or 3. Obviously, 1 should be abandoned. Thus, the factors of the perfect product should be 2 or 3.

The reason why we should use 3 as many as possible is

For 6, 3 * 3>2 * 2 * 2. Thus, the optimal product should contain no more than three 2.

Below is my accepted, O(N) solution.

class Solution {
    public:
    int integerBreak(int n) {
        if(n==2) return 1;
        if(n==3) return 2;
        int product = 1;
        while(n>4){
            product*=3;
            n-=3;
        }
        product*=n;
        
        return product;
    }
};