LeetCode-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.
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.
第一种解法也是我第一时间想到的解法:让area依次除area到sqrt(area),找到其中乘数相差最小的。
java代码:
public class Solution { public int[] constructRectangle(int area) { int[] arr=new int[]{area,1}; if(area<=0){ arr[0]=0; arr[1]=0; return arr; } for(int i=area;i>=Math.sqrt(area);i--){ if(area%i==0){ if(i>=area/i&&(i-area)<(arr[0]-arr[1])){ arr[0]=i; arr[1]=area/i; } } } return arr; } }
下面介绍一种效率更高的方法,那就是area除以一个比sqrt(area)小一点点的数,如果商为零,那么就是乘数相差最小的,也是长大于宽的。
public class Solution { public int[] constructRectangle(int area) { int[] result = new int[2]; if(area == 0){ return result; } int a = (int)Math.sqrt(area); while(area%a != 0){ a--; } int b = area/a; result[0] = b; result[1] = a; return result; } }