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: 7Explanation: 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

Runtime:?1 ms, faster than?82.83%?of?Java?online submissions for?Minimum Time Visiting All Points.

Memory Usage:?44.1 MB, less than?13.86%?of?Java?online submissions for?Minimum Time Visiting All Points.

因?yàn)樯舷伦笥遥ㄐ钡?，都?s，所以只要看2個(gè)方向差異的最大值就可以了，以此遞推，就能算出來的。