Description
You are given an integer 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 <= 1000coordinates[i].length == 2-10^4 <= coordinates[i][0], coordinates[i][1] <= 10^4coordinates contains no duplicate point.
Solutions
Solution 1: Mathematics
The time complexity is O(n), where n is the length of the coordinates array. The space complexity is O(1).
PythonJavaC++Go
class Solution:
def checkStraightLine(self, coordinates: List[List[int]]) -> bool:
x1, y1 = coordinates[0]
x2, y2 = coordinates[1]
for x, y in coordinates[2:]:
if (x - x1) * (y2 - y1) != (y - y1) * (x2 - x1):
return False
return True(code-box)
class Solution {
public boolean checkStraightLine(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.length; ++i) {
int x = coordinates[i][0], y = coordinates[i][1];
if ((x - x1) * (y2 - y1) != (y - y1) * (x2 - x1)) {
return false;
}
}
return true;
}
}(code-box)
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 x = coordinates[i][0], y = coordinates[i][1];
if ((x - x1) * (y2 - y1) != (y - y1) * (x2 - x1)) {
return false;
}
}
return true;
}
};(code-box)
func checkStraightLine(coordinates [][]int) bool {
x1, y1 := coordinates[0][0], coordinates[0][1]
x2, y2 := coordinates[1][0], coordinates[1][1]
for i := 2; i < len(coordinates); i++ {
x, y := coordinates[i][0], coordinates[i][1]
if (x-x1)*(y2-y1) != (y-y1)*(x2-x1) {
return false
}
}
return true
}(code-box)
