Leetcode 492 - Construct the Rectangle

https://leetcode.com/problems/construct-the-rectangle/

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.

https://www.cnblogs.com/grandyang/p/6390311.html
这道题让我们根据面积来求出矩形的长和宽，要求长和宽的差距尽量的小，那么就是说越接近正方形越好。那么我们肯定是先来判断一下是不是正方行，对面积开方，如果得到的不是整数，说明不是正方形。那么我们取最近的一个整数，看此时能不能整除，如果不行，就自减1，再看能否整除。最坏的情况就是面积是质数，最后减到了1，那么返回结果即可

    vector<int> constructRectangle(int area) {
        int r = sqrt(area);
        while (area % r != 0) --r;
        return {area / r, r};
    }

如果我们不想用开方运算sqrt的话，那就从1开始，看能不能整除，循环的终止条件是看平方值是否小于等于面积

    vector<int> constructRectangle(int area) {
        int r = 1;
        for (int i = 1; i * i <= area; ++i) {
            if (area % i == 0) r = i;
        }
        return {area / r, r};
    }

https://leetcode.com/problems/construct-the-rectangle/discuss/97210/3-line-Clean-and-easy-understand-solution

The W is always less than or equal to the square root of area
so we start searching at sqrt(area) till we find the result

public int[] constructRectangle(int area) {
        int w = (int)Math.sqrt(area);
 while (area%w!=0) w--;
 return new int[]{area/w, w};
}

http://www.wonter.net/archives/1102.html
比较简单的题目，容易想到应该让长和宽尽量接近$\sqrt{n}$即可
所以可以枚举宽，从1到$\sqrt{n}$取最接近$\sqrt{n}$的宽，那么长就可以直接算出来了

1. vector<int> constructRectangle(int area)
2. {
3. int l, w;
4. int t = sqrt(area);
5. for(int i = t; i >= 1; --i)
6. {
7. if(area % i == 0)
8. {
9. w = i;
10. break;
11. }
12. }
13. l = area / w;
14. return vector<int>{l, w};
15. }