给定N个坐标点的数组arr [] ，任务是检查是否可以使用这些坐标点形成一条直线。
Input: arr[] = {{0, 0}, {1, 1}, {2, 2}}
Output: Yes
Explanation:
Slope of every two points is same. That is 1.
Therefore, a straight line can be formed using these points.
Input: arr[] = {{0, 1}, {2, 0}}
Output: Yes
Explanation:
Two points in co-ordinate system always forms a straight line.
方法：想法是找到阵列中每对点之间的直线斜率，如果每对点的斜率相同，则这些点共同形成一条直线。
// Slope of line formed by
// two points (y2, y1), (x2, x1)
Slope of Line = y2 - y1
---------
x2 - x1
下面是上述方法的实现：
C++
// C++ implementation to check
// if a straight line
// can be formed using N points
#include
using namespace std;
// Function to check if a straight line
// can be formed using N points
bool isStraightLinePossible(
vector > arr, int n)
{
// First pair of point (x0, y0)
int x0 = arr[0].first;
int y0 = arr[0].second;
// Second pair of point (x1, y1)
int x1 = arr[1].first;
int y1 = arr[1].second;
int dx = x1 - x0, dy = y1 - y0;
// Loop to iterate over the points
for (int i = 0; i < n; i++) {
int x = arr[i].first, y = arr[i].second;
if (dx * (y - y1) != dy * (x - x1)){
cout << "NO";
return false;
}
}
cout << "YES";
return true;
}
// Driver Code
int main()
{
// Array of points
vector > arr =
{ { 0, 0 }, { 1, 1 }, { 3, 3 }, { 2, 2 } };
int n = 4;
// Function Call
isStraightLinePossible(arr, n);
return 0;
} Java
// Java implementation to check
// if a straight line can be
// formed using N points
import java.util.*;
class GFG{
static class pair
{
int first, second;
pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// Function to check if a straight line
// can be formed using N points
static void isStraightLinePossible(
ArrayList arr, int n)
{
// First pair of point (x0, y0)
int x0 = arr.get(0).first;
int y0 = arr.get(0).second;
// Second pair of point (x1, y1)
int x1 = arr.get(1).first;
int y1 = arr.get(1).second;
int dx = x1 - x0, dy = y1 - y0;
// Loop to iterate over the points
for(int i = 0; i < n; i++)
{
int x = arr.get(i).first;
int y = arr.get(i).second;
if (dx * (y - y1) != dy * (x - x1))
{
System.out.println("NO");
}
}
System.out.println("YES");
}
// Driver code
public static void main(String[] args)
{
// Array of points
ArrayList arr = new ArrayList<>();
arr.add(new pair(0, 0));
arr.add(new pair(1, 1));
arr.add(new pair(3, 3));
arr.add(new pair(2, 2));
int n = 4;
// Function Call
isStraightLinePossible(arr, n);
}
}
// This code is contributed by offbeat Python3
# Python3 implementation to check
# if a straight line can be formed
# using N points
# Function to check if a straight line
# can be formed using N points
def isStraightLinePossible(arr, n):
# First pair of point (x0, y0)
x0 = arr[0][0]
y0 = arr[0][1]
# Second pair of point (x1, y1)
x1 = arr[1][0]
y1 = arr[1][1]
dx = x1 - x0
dy = y1 - y0
# Loop to iterate over the points
for i in range(n):
x = arr[i][0]
y = arr[i][1]
if (dx * (y - y1) != dy * (x - x1)):
print("NO", end = "")
return False
print("YES", end = "")
return True
# Driver code
# Array of points
arr = [ [ 0, 0 ], [ 1, 1 ],
[ 3, 3 ], [ 2, 2 ] ]
n = 4
# Function Call
isStraightLinePossible(arr, n)
# This code is contributed by divyeshrabadiya07C#
// C# implementation to check
// if a straight line can be
// formed using N points
using System;
using System.Collections.Generic;
class GFG{
public class pair
{
public int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// Function to check if a straight line
// can be formed using N points
static void isStraightLinePossible(
List arr, int n)
{
// First pair of point (x0, y0)
int x0 = arr[0].first;
int y0 = arr[0].second;
// Second pair of point (x1, y1)
int x1 = arr[1].first;
int y1 = arr[1].second;
int dx = x1 - x0, dy = y1 - y0;
// Loop to iterate over the points
for(int i = 0; i < n; i++)
{
int x = arr[i].first;
int y = arr[i].second;
if (dx * (y - y1) != dy * (x - x1))
{
Console.WriteLine("NO");
}
}
Console.WriteLine("YES");
}
// Driver code
public static void Main(String[] args)
{
// Array of points
List arr = new List();
arr.Add(new pair(0, 0));
arr.Add(new pair(1, 1));
arr.Add(new pair(3, 3));
arr.Add(new pair(2, 2));
int n = 4;
// Function call
isStraightLinePossible(arr, n);
}
}
// This code is contributed by Princi Singh 输出：
YES
时间复杂度： O(N)