week10- Dynamic Programming-NO.343. Integer Break

最新推荐文章于 2023-11-20 23:11:11 发布

jingyu1621

最新推荐文章于 2023-11-20 23:11:11 发布

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探讨如何将一个整数拆分成多个正整数之和，以使这些数的乘积达到最大值。适用于大于等于2且小于等于58的整数。策略核心在于尽可能多地拆分出3。

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题目

Total Accepted: 39857
Total Submissions: 87649
Difficulty: Medium
Contributor: LeetCode

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.

https://leetcode.com/problems/integer-break/#/description

思路

题目需要将一个整数拆分成若干个数的和，使得这些数的乘积最大

通过观察发现，对于一个大于4的数，要使得乘积最大，需要将其拆分成尽可能多的3的和，最后剩下的数为2或4即可。例如10 = 3 + 3 + 4,14 = 3 + 3 + 3 + 3 + 2。

源程序

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