[LeetCode] 492. Construct the Rectangle

[grandyang]

 

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.

 

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，参见代码如下：

 

解法一：

class Solution {
public:
    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，参见代码如下：

 

解法二：

class Solution {
public:
    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};
    }
};

 

参考资料：

https://discuss.leetcode.com/topic/76469/3ms-concise-c

https://discuss.leetcode.com/topic/76314/3-line-clean-and-easy-understand-solution

 

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