492. Construction rectangle

1. Title Description

As a web developer , It's important to know how to plan the size of a page . therefore , Now give a specific rectangular page area , Your task is to design a length of L And the width is W A rectangular page that meets the following requirements . requirement ：
① The rectangular page you design must be equal to the given target area .
② Width W Should not be greater than length L , In other words , requirement L >= W .
③ length L Width and width W The gap should be as small as possible .
Return to one Array [L, W], among L and W Is the length and width of the web pages you design in order .

Input : 4
Output : [2, 2]
explain : The target area is 4, All possible construction schemes are [1,4], [2,2], [4,1].
But on request 2,[1,4] Unqualified ; According to the requirements 3,[2,2] Than [4,1] Better meet the requirements . So the output length L by 2, Width W by 2.

2. Their thinking

The key is length L Width and width W The gap should be as small as possible , Obviously L=W Minimum time difference , So from sqrt(area) Start looking left , The first thing that can be divided by area is what you want W.

• python3

class Solution:
    def constructRectangle(self, area: int) -> List[int]:
        wide = int(sqrt(area))
        while area % wide != 0:
            wide -= 1
        return [int(area/wide), wide]

• c++

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