给定一个数组nrr [] ，该数组arr []由代表N个点的坐标的N对整数组成，任务是找到由平行于给定点集的X和Y轴绘制的直线形成的最大矩形的面积。
例子：
Input: arr[] = {{0, 0}, {1, 1}}
Output: 1
Explanation: The area of the largest rectangle is 1 formed by the coordinates (0, 0), (0, 1), (1, 0), (1, 1).
Input: arr[] = {{-2, 0}, {2, 0}, {4, 0}, {4, 2}}
Output: 8
Explanation: The area of the largest rectangle possible is 8 ( length = 4 and breadth = 2 ) by the coordinates (-2, 0), (2, 0), (2, 2), (-2, 2).
方法：可以使用排序技术解决问题。请按照以下步骤解决问题：
- 将X和Y坐标存储在两个不同的数组中，例如x []和y [] 。
- 对数组x []和y []进行排序。
- 找到两个数组的最大相邻差并将其存储在变量X_Max和Y_Max中。
- 可能的最大矩形区域是X_Max和Y_Max的乘积。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to return the area of the largest
// rectangle formed by lines parallel to X
// and Y axis from given set of points
int maxRectangle(vector > sequence,
int size)
{
// Initialize two arrays
long long int X_Cord[size], Y_Cord[size];
// Store x and y coordinates
for (int i = 0; i < size; i++) {
X_Cord[i] = sequence[i][0];
Y_Cord[i] = sequence[i][1];
}
// Sort arrays
sort(X_Cord, X_Cord + size);
sort(Y_Cord, Y_Cord + size);
// Initialize max differences
long long int X_Max = 0, Y_Max = 0;
// Find max adjacent differences
for (int i = 0; i < size - 1; i++) {
X_Max = max(X_Max, X_Cord[i + 1]
- X_Cord[i]);
Y_Max = max(Y_Max, Y_Cord[i + 1]
- Y_Cord[i]);
}
// Return answer
return X_Max * Y_Max;
}
// Driver Code
int main()
{
// Given points
vector > point
= { { -2, 0 }, { 2, 0 }, { 4, 0 }, { 4, 2 } };
// Total points
int n = point.size();
// Function call
cout << maxRectangle(point, n);
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to return the area of the largest
// rectangle formed by lines parallel to X
// and Y axis from given set of points
static int maxRectangle(int[][] sequence,
int size)
{
// Initialize two arrays
int[] X_Cord = new int[size];
int[] Y_Cord = new int[size];
// Store x and y coordinates
for(int i = 0; i < size; i++)
{
X_Cord[i] = sequence[i][0];
Y_Cord[i] = sequence[i][1];
}
// Sort arrays
Arrays.sort(X_Cord);
Arrays.sort(Y_Cord);
// Initialize max differences
int X_Max = 0, Y_Max = 0;
// Find max adjacent differences
for(int i = 0; i < size - 1; i++)
{
X_Max = Math.max(X_Max, X_Cord[i + 1] -
X_Cord[i]);
Y_Max = Math.max(Y_Max, Y_Cord[i + 1] -
Y_Cord[i]);
}
// Return answer
return X_Max * Y_Max;
}
// Driver Code
public static void main(String[] args)
{
// Given points
int[][] point = { { -2, 0 }, { 2, 0 },
{ 4, 0 }, { 4, 2 } };
// Total points
int n = point.length;
// Function call
System.out.print(maxRectangle(point, n));
}
}
// This code is contributed by shikhasingrajputPython3
# Python3 program for the
# above approach
# Function to return the
# area of the largest
# rectangle formed by lines
# parallel to X and Y axis
# from given set of points
def maxRectangle(sequence, size):
# Initialize two arrays
X_Cord = [0] * size
Y_Cord = [0] * size
# Store x and y coordinates
for i in range(size):
X_Cord[i] = sequence[i][0]
Y_Cord[i] = sequence[i][1]
# Sort arrays
X_Cord.sort()
Y_Cord.sort()
# Initialize max differences
X_Max = 0
Y_Max = 0
# Find max adjacent differences
for i in range(size - 1):
X_Max = max(X_Max,
X_Cord[i + 1] -
X_Cord[i])
Y_Max = max(Y_Max,
Y_Cord[i + 1] -
Y_Cord[i])
# Return answer
return X_Max * Y_Max
# Driver Code
if __name__ == "__main__":
# Given points
point = [[-2, 0], [2, 0],
[4, 0], [4, 2]]
# Total points
n = len(point)
# Function call
print(maxRectangle(point, n))
# This code is contributed by ChitranayalC#
// C# program for the above approach
using System;
class GFG{
// Function to return the area of the largest
// rectangle formed by lines parallel to X
// and Y axis from given set of points
static int maxRectangle(int[,] sequence,
int size)
{
// Initialize two arrays
int[] X_Cord = new int[size];
int[] Y_Cord = new int[size];
// Store x and y coordinates
for(int i = 0; i < size; i++)
{
X_Cord[i] = sequence[i, 0];
Y_Cord[i] = sequence[i, 1];
}
// Sort arrays
Array.Sort(X_Cord);
Array.Sort(Y_Cord);
// Initialize max differences
int X_Max = 0, Y_Max = 0;
// Find max adjacent differences
for(int i = 0; i < size - 1; i++)
{
X_Max = Math.Max(X_Max, X_Cord[i + 1] -
X_Cord[i]);
Y_Max = Math.Max(Y_Max, Y_Cord[i + 1] -
Y_Cord[i]);
}
// Return answer
return X_Max * Y_Max;
}
// Driver Code
public static void Main(String[] args)
{
// Given points
int[,] point = { { -2, 0 }, { 2, 0 },
{ 4, 0 }, { 4, 2 } };
// Total points
int n = point.GetLength(0);
// Function call
Console.Write(maxRectangle(point, n));
}
}
// This code is contributed by shikhasingrajput输出：
8时间复杂度： O(N * log(N))，其中N是总点数。
辅助空间： O(N)