LeetCode 492. Construct the Rectangle（构造矩形）

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.

解法一

思路

这题就是说确定矩形的长和宽，使得面积与给定值相等，而且长宽尽可能接近。那么最优解就是长宽相等而且面积相等了，即x * x = area，但是这样的整数x却不一定存在。既然我们知道了x的最优取值，接下来就只需要将x逐步增大，直到出现area % x == 0，即存在整数x、y使得x * y = area就好了。

Python

class Solution(object):
    def constructRectangle(self, area):
        """
        :type area: int
        :rtype: List[int]
        """
        x = int(math.ceil(area ** 0.5))
        while area % x != 0:
            x += 1
        y = int(area / x)
        return [x, y]

Java

public class Solution {
    public int[] constructRectangle(int area) {
        int x = (int) Math.ceil(Math.sqrt(area));
        while (area % x != 0) {
            x++;
        }
        int y = area / x;
        return new int[] {x, y};
    }
}

C++

class Solution {
public:
    vector<int> constructRectangle(int area) {
        int x = (int) ceil(sqrt(area));
        while (area % x != 0) {
            x++;
        }
        int y = area / x;
        return vector<int> {x, y};
    }
};