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68 lines
2.3 KiB
Markdown
68 lines
2.3 KiB
Markdown
# [1725. Number Of Rectangles That Can Form The Largest Square](https://leetcode.com/problems/number-of-rectangles-that-can-form-the-largest-square/)
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## 题目
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You are given an array `rectangles` where `rectangles[i] = [li, wi]` represents the `ith` rectangle of length `li` and width `wi`.
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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`.
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Let `maxLen` be the side length of the **largest** square you can obtain from any of the given rectangles.
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Return *the **number** of rectangles that can make a square with a side length of* `maxLen`.
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**Example 1:**
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```
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Input: rectangles = [[5,8],[3,9],[5,12],[16,5]]
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Output: 3
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Explanation: The largest squares you can get from each rectangle are of lengths [5,3,5,5].
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The largest possible square is of length 5, and you can get it out of 3 rectangles.
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```
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**Example 2:**
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```
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Input: rectangles = [[2,3],[3,7],[4,3],[3,7]]
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Output: 3
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```
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**Constraints:**
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- `1 <= rectangles.length <= 1000`
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- `rectangles[i].length == 2`
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- `1 <= li, wi <= 10^9`
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- `li != wi`
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## 题目大意
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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 。
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## 解题思路
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- 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。
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## 代码
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```go
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package leetcode
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func countGoodRectangles(rectangles [][]int) int {
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minLength, count := 0, 0
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for i, _ := range rectangles {
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minSide := 0
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if rectangles[i][0] <= rectangles[i][1] {
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minSide = rectangles[i][0]
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} else {
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minSide = rectangles[i][1]
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}
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if minSide > minLength {
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minLength = minSide
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count = 1
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} else if minSide == minLength {
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count++
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}
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}
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return count
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}
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``` |