(LeetCode) 343. Integer Break

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探讨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。通过算法实现，针对特定输入值提供最优解。

343. Integer Break

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.

代码

class Solution {
public:
    int integerBreak(int n) {
        if(n==2) return 1;
        if(n==3) return 2;

        int res = 1;
        while (n>4){
            res *= 3;
            n -= 3;
        }

        return res*n;

    }
};