zoukankan      html  css  js  c++  java
  • 492. 构造矩形 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: 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. 
    给定矩形的面积area,返回矩形的长度L和宽度W,使得L和W的差值最小。
    1. public class Solution {
    2. public int[] ConstructRectangle(int area) {
    3. if (area == 0) {
    4. int[] arr = { };
    5. return arr;
    6. }
    7. int[] result = { area, 1 };
    8. for(int height = 1; height <= area; height++) {
    9. if (area % height == 0) {
    10. int width = area / height;
    11. if (width < height) {
    12. break;
    13. }else if (width - height < result[0] - result[1]) {
    14. result[0] = width;
    15. result[1] = height;
    16. }
    17. }
    18. }
    19. return result;
    20. }
    21. }





  • 相关阅读:
    python库--pandas--文本文件读取
    python库--flashtext--大规模数据清洗利器
    PyCharm--帮助文档
    Git--命令
    symfony doctrine generate entity repository
    [转]MySQL性能优化的最佳20+条经验
    svn使用
    一致性hash
    JavaScript学习笔记 1
    curl发出请求
  • 原文地址:https://www.cnblogs.com/xiejunzhao/p/314a8db4895430ed533557f4e7ade523.html
Copyright © 2011-2022 走看看