492. 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: 4Output: [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.

public class Solution {    public int[] constructRectangle(int area) {        int[] res=new int[2];        int small=area-1;        int a,b;        for(int i=1;i<=area;i++)        {            if(area%i==0)            {                a=area/i;                b=i;                if(a>=b&&(a-b)<=small)                {                    small=a-b;                    res[0]=a;                    res[1]=b;                }            }        }        return res;    }}

改进：先求area的开方，如果开方值是整数，return的就是开方值，因为差距最小；如果开方值不是整数，判断area能否整除int型的开方值，如果不能，开方值-1，直到找到可以整除的情况

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};         }  }