leetcode1266. Minimum Time Visiting All Points

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.

因为上下左右，包括斜的，都是1s，所以只要看2个方向差异的最大值就可以了，以此递推，就能算出来的。