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 1:

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

Example 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.

这道题给了一堆二维坐标上的点，问这些点是否在一条直线上。初中的时候应该就学过两点确定一条直线，同时还有如何判断三点共线的问题，本质上就是判断三点中任意两个点组成的直线的斜率是否相同。计算两个点组成的直线的斜率，就是二者的纵坐标之差除以横坐标之差，但用除法的话就存在一个除数为0的问题，所以在比较两条直线的斜率是否相等时，将其变为乘法的形式，能有效的避免除数为0的情况。这道题说了给定的点的个数至少有两个，则可以用前两个点来确定一条直线，然后从第三个点开始遍历，判断每个遍历到的点和前两个点是否共线，就是不停的在验证三点共线问题，若某个点不在直线上，则返回 false，否则最后返回 true 即可，参见代码如下：

class Solution {
public:
    bool checkStraightLine(vector<vector<int>>& coordinates) {
        int x1 = coordinates[0][0], y1 = coordinates[0][1];
        int x2 = coordinates[1][0], y2 = coordinates[1][1];
        for (int i = 2; i < coordinates.size(); ++i) {
            int x3 = coordinates[i][0], y3 = coordinates[i][1];
            if ((x2 - x1) * (y3 - y1) != (y2 - y1) * (x3 - x1)) return false;
        }
        return true;
    }
};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/1232

参考资料：

https://leetcode.com/problems/check-if-it-is-a-straight-line/

https://leetcode.com/problems/check-if-it-is-a-straight-line/discuss/408984/JavaPython-3-check-slopes-short-code-w-explanation-and-analysis.

https://leetcode.com/problems/check-if-it-is-a-straight-line/discuss/620096/Java-Python3-CPP-or-Simple-code-with-explanation-or-100-fast-or-O(1)-space

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