You can move according to these rules:
In 1 second, you can either:
move vertically by one unit,
move horizontally by one unit, or
move diagonally sqrt(2) units (in other words, move one unit vertically then one unit horizontally in 1 second).
You have to visit the points in the same order as they appear in the array.
You are allowed to pass through points that appear later in the order, but these do not count as visits.
Example 1:
Input: points = [[1,1],[3,4],[-1,0]]
Output: 7
Explanation: One optimal path is
[1,1] -> [2,2] -> [3,3] -> [3,4] -> [2,3] -> [1,2] -> [0,1] -> [-1,0]
Time from [1,1] to [3,4] = 3 seconds
Time from [3,4] to [-1,0] = 4 seconds
Total time = 7 seconds
Example 2:
Input: points = [[3,2],[-2,2]]
Output: 5
Constraints:
points.length == n
1 <= n <= 100
points[i].length == 2
-1000 <= points[i][0], points[i][1] <= 1000
class Solution:
def minTimeToVisitAllPoints(self, points: List[List[int]]) -> int:
step = 0
for i in range(len(points) - 1):
step += max(abs(points[i][0] - points[i+1][0]), abs(points[i][1] - points[i+1][1]))
return step
문제를 막상 봤을 때 복잡하게 생각하기 쉽지만, 대각선으로 가는 것도 1스텝으로 치기 때문에 갈 수 있는 부분까지 최대한 대각선(또는 1 혹은 -1의 기울기)으로 간 다음, 가로나 세로를 가는 것이 점과 점 사이를 최소한의 스텝으로 움직이는 방법이다.