Topic
- Array
- Geometry
Description
https://leetcode.com/problems/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 in1
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: 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
观察并得出过滤。
Submission
public class MinimumTimeVisitingAllPoints {
public int minTimeToVisitAllPoints(int[][] points) {
int sum = 0;
for(int i = 1; i < points.length; i++) {
sum += (Math.max(Math.abs(points[i][0] - points[i - 1][0]), //
Math.abs(points[i][1] - points[i - 1][1])));
}
return sum;
}
}
Test
import static org.junit.Assert.*;
import org.junit.Test;
public class MinimumTimeVisitingAllPointsTest {
@Test
public void test() {
MinimumTimeVisitingAllPoints obj = new MinimumTimeVisitingAllPoints();
assertEquals(7, obj.minTimeToVisitAllPoints(new int[][] {{1, 1}, {3, 4}, {-1, 0}}));
assertEquals(5, obj.minTimeToVisitAllPoints(new int[][] {{3, 2},{-2, 2}}));
}
}