leetcode-492-Construct the Rectangle

Posted 2020-11-01 chenjx85

题目描述：

For a web developer, it is very important to know how to design a web page‘s size. So, given a specific rectangular web page’s area, your job by now is to design a rectangular web page, whose length L and width W satisfy the following requirements:

1. The area of the rectangular web page you designed must equal to the given target area.

2. The width W should not be larger than the length L, which means L >= W.

3. The difference between length L and width W should be as small as possible.

You need to output the length L and the width W of the web page you designed in sequence.

 

Example:

Input: 4
Output: [2, 2]
Explanation: The target area is 4, and all the possible ways to construct it are [1,4], [2,2], [4,1]. 
But according to requirement 2, [1,4] is illegal; according to requirement 3,  [4,1] is not optimal compared to [2,2]. So the length L is 2, and the width W is 2.

 

Note:

1. The given area won‘t exceed 10,000,000 and is a positive integer
2. The web page‘s width and length you designed must be positive integers.

 

要完成的函数：

vector<int> constructRectangle(int area) 

 

说明：

1、这道题目不难，给定一个乘积，要求输出两个因子（大的在前，小的在后），两个因子的差应该越小越好。

2、看到这些限定条件，我们首选就是开方，然后在开方得到的数值附近找到一个乘积能够整除的数。

代码如下：

    vector<int> constructRectangle(int area) 
    {
        int eu=ceil(sqrt(area));
        while(area%eu)
            eu+=1;
        return {eu,area/eu};
    }

上述代码accepted，实测80ms，beats 27.85% of cpp submissions……

这么低的吗？但我看评论区的代码采用的也是跟我一样的方法，为什么他们就能4ms……

 

3、改进：

细细对比了一下评论区的代码，他们采用的是eu不断地减一，而我用的是eu不断地加一。

因为eu小的话比较容易除？比如2889除以11，跟2889除以3比起来，肯定后者比较好做除法。

所以如果采用eu不断地减一的方法，的确能够节省很多做除法的时间，而两者的效果是一样的。

代码如下：

    vector<int> constructRectangle(int area) 
    {
        int eu=floor(sqrt(area));
        while(area%eu)
            eu-=1;
        return {area/eu,eu};
    }

实测3ms，beats 98.48% of cpp submissions。