1725. Number Of Rectangles That Can Form The Largest Square #

题目 #

You are given an array rectangles where rectangles[i] = [li, wi] represents the ith rectangle of length li and width wi.

You can cut the ith rectangle to form a square with a side length of k if both k <= li and k <= wi. For example, if you have a rectangle [4,6], you can cut it to get a square with a side length of at most 4.

Let maxLen be the side length of the largest square you can obtain from any of the given rectangles.

Return the number of rectangles that can make a square with a side length of maxLen.

Example 1:

Input: rectangles = [[5,8],[3,9],[5,12],[16,5]]
Output: 3
Explanation: The largest squares you can get from each rectangle are of lengths [5,3,5,5].
The largest possible square is of length 5, and you can get it out of 3 rectangles.

Example 2:

Input: rectangles = [[2,3],[3,7],[4,3],[3,7]]
Output: 3

Constraints:

• 1 <= rectangles.length <= 1000
• rectangles[i].length == 2
• 1 <= li, wi <= 10^9
• li != wi

题目大意 #

给你一个数组 rectangles ，其中 rectangles[i] = [li, wi] 表示第 i 个矩形的长度为 li 、宽度为 wi 。如果存在 k 同时满足 k <= li 和 k <= wi ，就可以将第 i 个矩形切成边长为 k 的正方形。例如，矩形 [4,6] 可以切成边长最大为 4 的正方形。设 maxLen 为可以从矩形数组 rectangles 切分得到的 最大正方形 的边长。返回可以切出边长为 maxLen 的正方形的矩形 数目 。

解题思路 #

• 简单题。扫描数组中的每一个矩形，先找到边长较小的那条边，作为正方形的边长。扫描过程中动态更新最大的正方形边长，并累加计数。循环一遍结束，输出最终计数值即可。

代码 #

package leetcode

func countGoodRectangles(rectangles [][]int) int {
	minLength, count := 0, 0
	for i, _ := range rectangles {
		minSide := 0
		if rectangles[i][0] <= rectangles[i][1] {
			minSide = rectangles[i][0]
		} else {
			minSide = rectangles[i][1]
		}
		if minSide > minLength {
			minLength = minSide
			count = 1
		} else if minSide == minLength {
			count++
		}
	}
	return count
}