1266. Minimum Time Visiting All Points-访问所有点的最小时间

On a 2D plane, there are n points with integer coordinates points[i] = [xi, yi]. Return the minimum time in seconds to visit all the points in the order given by points.

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

Solution in python:

class Solution:
    def minTimeToVisitAllPoints(self, points: List[List[int]]) -> int:
        count = 0
        origin = points[0]
        i = 1
        while i < len(points):
            if points[i][0] == origin[0]:
                count += abs(points[i][1]-origin[1])
            elif points[i][1] == origin[1]:
                count += abs(points[i][0]-origin[0])
            else:
                dx = abs(points[i][0]-origin[0])
                dy = abs(points[i][1]-origin[1])
                count += min(dx, dy) + abs(dx-dy)
            origin = points[i]
            i += 1
        return count