1232 Check If It Is a Straight Line 缀点成线

Description:
You are given an array coordinates, coordinates[i] = [x, y], where [x, y] represents the coordinate of a point. Check if these points make a straight line in the XY plane.

Example:

Example 1:

coordinates 1

Input: coordinates = [[1,2],[2,3],[3,4],[4,5],[5,6],[6,7]]
Output: true

Example 2:

coordinates 2

Input: coordinates = [[1,1],[2,2],[3,4],[4,5],[5,6],[7,7]]
Output: false

Constraints:

2 <= coordinates.length <= 1000
coordinates[i].length == 2
-10^4 <= coordinates[i][0], coordinates[i][1] <= 10^4
coordinates contains no duplicate point.

题目描述:
在一个 XY 坐标系中有一些点，我们用数组 coordinates 来分别记录它们的坐标，其中 coordinates[i] = [x, y] 表示横坐标为 x、纵坐标为 y 的点。

请你来判断，这些点是否在该坐标系中属于同一条直线上，是则返回 true，否则请返回 false。

示例 :

示例 1：

坐标系1

输入：coordinates = [[1,2],[2,3],[3,4],[4,5],[5,6],[6,7]]
输出：true

示例 2：

坐标系2

输入：coordinates = [[1,1],[2,2],[3,4],[4,5],[5,6],[7,7]]
输出：false

提示：

2 <= coordinates.length <= 1000
coordinates[i].length == 2
-10^4 <= coordinates[i][0], coordinates[i][1] <= 10^4
coordinates 中不含重复的点

思路:

参考LeetCode #1037 Valid Boomerang 有效的回旋镖
时间复杂度O(n), 空间复杂度O(1)

代码:
C++:

class Solution 
{
public:
    bool checkStraightLine(vector<vector<int>>& coordinates) 
    {
        int x = coordinates[1][0] - coordinates[0][0], y = coordinates[1][1] - coordinates[0][1];
        for (int i = 1; i < coordinates.size(); i++) if ((coordinates[i][0] - coordinates[0][0]) * y != x * (coordinates[i][1] - coordinates[0][1])) return false;
        return true;
    }
};

Java:

class Solution {
    public boolean checkStraightLine(int[][] coordinates) {
        int x = coordinates[1][0] - coordinates[0][0], y = coordinates[1][1] - coordinates[0][1];
        for (int i = 1; i < coordinates.length; i++) if ((coordinates[i][0] - coordinates[0][0]) * y != x * (coordinates[i][1] - coordinates[0][1])) return false;
        return true;
    }
}

Python:

class Solution:
    def checkStraightLine(self, coordinates: List[List[int]]) -> bool:
        return all((coordinates[1][1] - coordinates[0][1]) * (coordinates[i][0] - coordinates[0][0]) == (coordinates[i][1] - coordinates[0][1]) * (coordinates[1][0] - coordinates[0][0]) for i in range(2, len(coordinates)))