LeetCode 343. Integer Break 题解

原创于 2016-10-28 17:01:52 发布 · 281 阅读

0 ·

CC 4.0 BY-SA版权

leetcode 专栏收录该内容

38 篇文章

订阅专栏

探讨如何通过算法将给定的正整数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，对于整数10，最优解为3+3+4，其乘积为36。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

343. Integer Break

My Submissions

Total Accepted: 29402
Total Submissions: 66455
Difficulty: Medium
Contributors: Admin

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.

解题思路:

将n分解为尽可能多的3的和.原因为pow(2,3)<pow(3,2)且2*3=3*2。

代码展示：

class Solution {
public:
    int integerBreak(int n) {
        if(n==2) return 1;
        if(n==3) return 2;
        vector<int> res;
        while(n-3>=2)
        {
            res.push_back(3);
            n-=3;
        }
        if(n) res.push_back(n);
        int ans=1;
        for(int i=0;i<res.size();i++)
        {
            ans*=res[i];
        }
        return ans;
        
    }
};