public class Solution extends Object
1266 - Minimum Time Visiting All Points.
Easy
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:
1 second, you can either:
sqrt(2) units (in other words, move one unit vertically then one unit horizontally in 1 second).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 == n1 <= n <= 100points[i].length == 2-1000 <= points[i][0], points[i][1] <= 1000| Constructor and Description |
|---|
Solution() |
| Modifier and Type | Method and Description |
|---|---|
int |
minTimeToVisitAllPoints(int[][] points) |
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