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.

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.

public class Solution {public int[] constructRectangle(int area) {if (area < 0)return null;int[] a = new int[2];if (area == 0) {a[0] = 0;a[1] = 0;return a;}int mid = (int)Math.sqrt(area);if (mid == 1) {a[0] = area;a[1] = 1;return a;}int l1 = 0, l2 = 0;int w1 = 0, w2 = 0;int sub1 = 0, sub2 = 0;//area是平方数的时候if (mid * mid == area) {a[0] = mid;a[1] = mid;return a;} else {for (int i = mid; i > 0; --i) {if (area % i == 0) {l1 = area / i;w1 = i;sub1 = mid - i;break;}}for (int j = mid; j <= area; ++j) {if (area % j == 0) {w2 = area / j;l2 = j;sub2 = j - mid;}}}if (sub1 <= sub2) {a[0] = l1;a[1] = w1;} else {a[0] = l2;a[1] = w2;}return a;}
}