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