LeetCode之Construct the Rectangle
【摘要】 1、题目
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 ...
1、题目
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:
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1. The area of the rectangular web page you designed must equal to the given target area.
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2. The width W should not be larger than the length L, which means L >= W.
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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:
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Input: 4
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Output: [2, 2]
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Explanation: The target area is 4, and all the possible ways to construct it are [1,4], [2,2], [4,1].
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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.
2、代码实现
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public class Solution {
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public int[] constructRectangle(int area) {
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if (area < 0)
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return null;
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int[] a = new int[2];
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if (area == 0) {
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a[0] = 0;
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a[1] = 0;
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return a;
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}
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int mid = (int)Math.sqrt(area);
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if (mid == 1) {
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a[0] = area;
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a[1] = 1;
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return a;
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}
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int l1 = 0, l2 = 0;
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int w1 = 0, w2 = 0;
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int sub1 = 0, sub2 = 0;
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//area是平方数的时候
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if (mid * mid == area) {
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a[0] = mid;
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a[1] = mid;
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return a;
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} else {
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for (int i = mid; i > 0; --i) {
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if (area % i == 0) {
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l1 = area / i;
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w1 = i;
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sub1 = mid - i;
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break;
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}
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}
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for (int j = mid; j <= area; ++j) {
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if (area % j == 0) {
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w2 = area / j;
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l2 = j;
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sub2 = j - mid;
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}
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}
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}
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if (sub1 <= sub2) {
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a[0] = l1;
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a[1] = w1;
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} else {
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a[0] = l2;
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a[1] = w2;
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}
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return a;
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}
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}
文章来源: chenyu.blog.csdn.net,作者:chen.yu,版权归原作者所有,如需转载,请联系作者。
原文链接:chenyu.blog.csdn.net/article/details/69939276
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