leetcode1266. Minimum Time Visiting All Points

Posted 2020-11-22 seyjs

题目如下：

On a plane there are n points with integer coordinates points[i] = [xi, yi]. Your task is to find the minimum time in seconds to visit all points.

You can move according to the next rules:

• In one second always you can either move vertically, horizontally by one unit or diagonally (it means to move one unit vertically and one unit horizontally in one second).
• You have to visit the points in the same order as they appear in the array. 

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

解题思路：优先走斜线，直到满足起点的x坐标等于终点的x坐标或者起点的y坐标等于终点的y坐标；然后走直线，直至到达终点。

代码如下：

class Solution(object):
    def minTimeToVisitAllPoints(self, points):
        """
        :type points: List[List[int]]
        :rtype: int
        """
        def calc(x1,y1,x2,y2):
            distance = min(abs(x1-x2),abs(y1-y2))
            return distance + abs(abs(x1-x2) - distance) + abs(abs(y1-y2) - distance)
        res = 0
        for i in range(len(points)-1):
            res += calc(points[i][0],points[i][1],points[i+1][0],points[i+1][1])
        return res