492. Construct the Rectangle 构造矩形

@TOC

# 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.
``````

# 解题方法

# Java解法

``````public class Solution {
public int[] constructRectangle(int area) {
double sqrt = Math.sqrt(area);
int int_sqrt = (int) sqrt;
if(int_sqrt * int_sqrt == area){
}else{
for(int i= int_sqrt; i >= 1; i--){
if(area % i == 0){
break;//不要忘记
}
}
}
}
}
``````

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

# python解法

``````class Solution(object):
def constructRectangle(self, area):
"""
:type area: int
:rtype: List[int]
"""
sqrt = int(math.sqrt(area))
for w in range(sqrt, 0, -1):
if area % w == 0:
return [area / w, w]
return [area, 1]
``````

# 日期

2017 年 4 月 3 日 2018 年 11 月 15 日 —— 时间太快，不忍直视