343. Integer Break

最新推荐文章于 2022-04-12 11:06:46 发布

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探讨如何将正整数n拆分为至少两个正整数之和，并最大化这些整数的乘积。例如，当n=10时，最佳拆分方案为3+3+4，此时的最大乘积为36。

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.

class Solution {
public:
    int integerBreak(int n) {
        if(n < 4) return n-1;
        int res = 1;
        while(n > 2){//看n包含多少个3,把他们相乘,直到n<=2
            res *= 3;
            n -= 3;
        }
        if(n == 0) return res;//n可以整除3，res就是各个3相乘
        if(n == 1) return (res / 3 ) * 4;//除3余1，把其中的一个3加1变为4再相乘
        if(n == 2) return res * 2;//除3余2,则可直接把2与res相乘
    }
};