LeetCode-Construct the Rectangle

最新推荐文章于 2020-04-17 10:13:46 发布

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本文介绍了一种用于设计网页尺寸的算法，旨在根据给定的目标面积找到最接近正方形且满足长度大于等于宽度的长宽组合。通过遍历所有可能的长度，筛选出符合条件的长宽比，最终确定最优解。

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Description：
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:

The area of the rectangular web page you designed must equal to the given target area.
The width W should not be larger than the length L, which means L >= W.
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:

The given area won’t exceed 10,000,000 and is a positive integer
The web page’s width and length you designed must be positive integers.

题意：给定一个网页的矩形空间的面积target，计算最优的长L和宽W；需要满足三个条件：

满足L x W = target
L >= W
dif = |l_i - w_i|= argmin_i|l_i - w_i|

解法：穷举所有的可能，首先需要排除那些不符合条件1和2的，再来选择最优的那个；

Java

class Solution {
    public int[] constructRectangle(int area) {
        int L = 0;
        int W = 0;
        int dif = Integer.MAX_VALUE;
        for (int l = 1; l <= area; l++) {
            int w = area / l;
            if (l * w != area || l < w) continue;
            if (l - w < dif) {
                L = l;
                W = w;
                dif = l - w;
            } 
        }
        return new int[] {L, W};
    }
}