LeetCode 1266. Minimum Time Visiting All Points (Easy)

本文探讨了在一个由整数坐标构成的行星上，如何找到遍历所有n个点所需的最短时间。通过分析两种移动策略，文章详细解释了如何在一秒内垂直、水平或对角线移动一单位距离，以确保按给定点阵顺序访问每个点。举例说明了不同输入下最优路径的计算过程，同时提供了一种简单易实现的方法。

Description:
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:
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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

Analysis:
两种方法:

先走斜线，直到到达一个横坐标或纵坐标与终点相同的点，然后开始走直线。
先走直线，直到到达一个可以通过走斜线到达终点的点，然后开始走斜线。

两种方法等价，但是从编程角度出发，第一种方法更容易实现。

方法1具体步骤:

先走斜线，这一步走的距离 $diagDistance = abs(x_{start} - x_{end}) + abs(y_{end} - y_{end})$ 。
再走直线，需要注意的是，此时当前点与终点在x方向与y方向上的距离，相比于起点，都减少了 $d i a g D i s t a n c e$ 。所以这一步走的距离 $distance = (abs(x_{start} - x_{end}) - diagDistance) + (abs(y_{start} - y_{end}) - diagDistance)$ 。

Code:

class Solution {
    public int minTimeToVisitAllPoints(int[][] points) {
        int distances = 0;
        for(int i = 0; i < points.length-1; i++) {
            int distance = calc(points[i], points[i+1]);
            distances += distance;
        }
        
        return distances;
    }
    
    public int calc(int[] pointX, int[] pointY) {
        int diagDistance = Math.min(Math.abs(pointX[0] - pointY[0]), Math.abs(pointX[1] - pointY[1]));
        int distance = diagDistance + (Math.abs(pointX[0] - pointY[0]) - diagDistance) + (Math.abs(pointX[1] - pointY[1]) - diagDistance);
        
        return distance;
    }
}