1232. Check If It Is a Straight Line

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.

class Solution {
    public boolean checkStraightLine(int[][] coordinates) {
        double k = (double)(coordinates[1][1] - coordinates[0][1]) / (coordinates[1][0] - coordinates[0][0]);
        for(int i = 2; i < coordinates.length; i++){
            double kt = (double)(coordinates[i][1] - coordinates[i-1][1]) / (coordinates[i][0] - coordinates[i-1][0]);
            if(k != kt) return false;
        }
        return true;
    }
}

先想到bruteforce，比较每两个点的斜率，注意4/5在int的时候是0，要先(double) 4/5这样。然后比较就可.

class Solution {
    public boolean checkStraightLine(int[][] coordinates) {
        int t1 = coordinates[1][1] - coordinates[0][1];        
        int t2 = coordinates[1][0] - coordinates[0][0];
        
        for(int i = 2; i < coordinates.length; i++){
            if((coordinates[i][1] - coordinates[i-1][1]) * t2 != t1 * (coordinates[i][0] - coordinates[i-1][0])) return false;
        }
        return true;
    }
}

 然后这个方法用上面的公式，不求斜率直接比交叉的积，不相同就return false