给定一个二维数组矩形 [][] 表示维度为L * B的矩形 {(0, 0), (L, 0), (0, B), (L, B)} 和另一个二维数组的顶点, 笛卡尔坐标系中大小为N 的点[][] 。从给定阵列的每个点绘制一条平行于X 轴的水平线和一条平行于Y 轴的垂直线。任务是计算给定矩形中存在的所有可能的矩形。
例子：
Input: Rectangle[][] = {{0, 0}, {5, 0}, {0, 5}, {5, 5}}, points[][] ={{1, 2}, {3, 4}}
Output: 9
Explanation:
Draw a horizontal line and a vertical line at points{1, 2} and points{3,4}.
Therefore, the required output is 9.
Input: Rectangle[][] = {{0, 0}, {4, 0}, {0, 5}, {4, 5}}, points[][] = {{1, 3}, {2, 3}, {3, 3}}
Output: 12
方法：想法是遍历数组并计算通过给定点 [][] 数组的所有不同的水平线和垂直线。最后，返回(不同水平线的计数 – 1) * (垂直线的计数 – 1)的乘法值。请按照以下步骤解决问题：
- 创建两个集合，比如cntHor ， cntVer来存储所有通过points[][]数组的不同水平线和垂直线。
- 遍历 points[][] 数组。
- 使用cntHor 中的元素计数计算所有不同的水平线。
- 使用cntVer 中的元素计数计算所有不同的垂直线。
- 最后，打印(count of elements in cntHor – 1) * (count of elements in cntVer – 1) 的值。
下面是上述方法的实现：
C++
_ _ _ _ _
5|_|_ _|_ _|
4| | |3,4|
3|_|_ _|_ _|
2| |1,2| |
1|_|_ _|_ _|
0 1 2 3 4 5Java
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to get the count
// of ractangles
int cntRect(int points[][2], int N,
int rectangle[][2])
{
// Store distinct
// horizontal lines
unordered_set cntHor;
// Store distinct
// Vertical lines
unordered_set cntVer;
// Insert horizontal line
// passing through 0
cntHor.insert(0);
// Insert vertical line
// passing through 0.
cntVer.insert(0);
// Insert horizontal line
// passing through rectangle[3][0]
cntHor.insert(rectangle[3][0]);
// Insert vertical line
// passing through rectangle[3][1]
cntVer.insert(rectangle[3][1]);
// Insert all horizontal and
// vertical lines passing through
// the given array
for (int i = 0; i < N; i++) {
// Insert all horizontal lines
cntHor.insert(points[i][0]);
// Insert all vertical lines
cntVer.insert(points[i][1]);
}
return (cntHor.size() - 1) *
(cntVer.size() - 1);
}
// Driver Code
int main()
{
int rectangle[][2] = {{0, 0}, {0, 5},
{5, 0}, {5, 5}};
int points[][2] = {{1, 2}, {3, 4}};
int N = sizeof(points) / sizeof(points[0]);
cout< Python3
// Java program to implement
// the above approach
import java.io.*;
import java.util.*;
class GFG{
// Function to get the count
// of ractangles
public static int cntRect(int points[][], int N,
int rectangle[][])
{
// Store distinct
// horizontal lines
HashSet cntHor = new HashSet<>();
// Store distinct
// Vertical lines
HashSet cntVer = new HashSet<>();
// Insert horizontal line
// passing through 0
cntHor.add(0);
// Insert vertical line
// passing through 0.
cntVer.add(0);
// Insert horizontal line
// passing through rectangle[3][0]
cntHor.add(rectangle[3][0]);
// Insert vertical line
// passing through rectangle[3][1]
cntVer.add(rectangle[3][1]);
// Insert all horizontal and
// vertical lines passing through
// the given array
for(int i = 0; i < N; i++)
{
// Insert all horizontal lines
cntHor.add(points[i][0]);
// Insert all vertical lines
cntVer.add(points[i][1]);
}
return (cntHor.size() - 1) *
(cntVer.size() - 1);
}
// Driver Code
public static void main(String args[])
{
int rectangle[][] = { { 0, 0 }, { 0, 5 },
{ 5, 0 }, { 5, 5 } };
int points[][] = { { 1, 2 }, { 3, 4 } };
int N = points.length;
System.out.println(cntRect(points, N,
rectangle));
}
}
// This code is contributed by hemanth gadarla C#
# Python3 program to implement
# the above approach
# Function to get the count
# of ractangles
def cntRect(points, N,
rectangle):
# Store distinct
# horizontal lines
cntHor = set([])
# Store distinct
# Vertical lines
cntVer = set([])
# Insert horizontal line
# passing through 0
cntHor.add(0)
# Insert vertical line
# passing through 0.
cntVer.add(0)
# Insert horizontal line
# passing through rectangle[3][0]
cntHor.add(rectangle[3][0])
# Insert vertical line
# passing through rectangle[3][1]
cntVer.add(rectangle[3][1])
# Insert all horizontal and
# vertical lines passing through
# the given array
for i in range (N):
# Insert all horizontal lines
cntHor.add(points[i][0])
# Insert all vertical lines
cntVer.add(points[i][1])
return ((len(cntHor) - 1) *
(len(cntVer) - 1))
# Driver Code
if __name__ == "__main__":
rectangle = [[0, 0], [0, 5],
[5, 0], [5, 5]]
points = [[1, 2], [3, 4]]
N = len(points)
print (cntRect(points, N, rectangle))
# This code is contributed by ChitranayalJavascript
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to get the count
// of ractangles
public static int cntRect(int [,]points,
int N, int [,]rectangle)
{
// Store distinct
// horizontal lines
HashSet cntHor = new HashSet();
// Store distinct
// Vertical lines
HashSet cntVer = new HashSet();
// Insert horizontal line
// passing through 0
cntHor.Add(0);
// Insert vertical line
// passing through 0.
cntVer.Add(0);
// Insert horizontal line
// passing through rectangle[3,0]
cntHor.Add(rectangle[3, 0]);
// Insert vertical line
// passing through rectangle[3,1]
cntVer.Add(rectangle[3, 1]);
// Insert all horizontal and
// vertical lines passing through
// the given array
for(int i = 0; i < N; i++)
{
// Insert all horizontal lines
cntHor.Add(points[i, 0]);
// Insert all vertical lines
cntVer.Add(points[i, 1]);
}
return (cntHor.Count - 1) *
(cntVer.Count - 1);
}
// Driver Code
public static void Main(String []args)
{
int [,]rectangle = {{0, 0}, {0, 5},
{5, 0}, {5, 5}};
int [,]points = {{1, 2}, {3, 4}};
int N = points.GetLength(0);
Console.WriteLine(cntRect(points, N,
rectangle));
}
}
// This code is contributed by 29AjayKumar 输出：
时间复杂度： O(N)
辅助空间： O(N)
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