LeetCode:Integer Break

最新推荐文章于 2019-04-07 02:16:37 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。通过动态规划的方法，实现了寻找给定整数n所能达到的最大乘积。

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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 == 2)
        return 1;
        vector<int> breakProduct = {1, 2};
        for(int i = 3; i <= n; i ++)
        {
            int maxPro = i - 1;
            int pro1;
            int pro2;
            for(int j = 1; j < i - 1; j ++)
            {
                pro1 = j * breakProduct[i - j - 1];
                if(pro1 > maxPro)
                    maxPro = pro1;
                pro2 = j * (i - j);
                if(pro2 > maxPro)
                    maxPro = pro2;
            }
            breakProduct.push_back(maxPro);
        }
        return breakProduct[n - 1];
    }
};