LeetCode - Easy - 1266. Minimum Time Visiting All Points

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 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: 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

觀察並得出過濾。

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}}));
	}
}