java.lang.Object
g1201_1300.s1266_minimum_time_visiting_all_points.Solution

public class Solution extends Object
1266 - Minimum Time Visiting All Points.<p>Easy</p> <p>On a 2D plane, there are <code>n</code> points with integer coordinates <code>points[i] = [x<sub>i</sub>, y<sub>i</sub>]</code>. Return <em>the <strong>minimum time</strong> in seconds to visit all the points in the order given by</em> <code>points</code>.</p> <p>You can move according to these rules:</p> <ul> <li>In <code>1</code> second, you can either: <ul> <li>move vertically by one unit,</li> <li>move horizontally by one unit, or</li> <li>move diagonally <code>sqrt(2)</code> units (in other words, move one unit vertically then one unit horizontally in <code>1</code> second).</li> </ul> </li> <li>You have to visit the points in the same order as they appear in the array.</li> <li>You are allowed to pass through points that appear later in the order, but these do not count as visits.</li> </ul> <p><strong>Example 1:</strong></p> <p><img src="https://assets.leetcode.com/uploads/2019/11/14/1626_example_1.PNG" alt="" /></p> <p><strong>Input:</strong> points = [[1,1],[3,4],[-1,0]]</p> <p><strong>Output:</strong> 7</p> <p><strong>Explanation:</strong> One optimal path is <strong>[1,1]</strong> -> [2,2] -> [3,3] -> <strong>[3,4]</strong> -> [2,3] -> [1,2] -> [0,1] -> <strong>[-1,0]</strong></p> <p>Time from [1,1] to [3,4] = 3 seconds</p> <p>Time from [3,4] to [-1,0] = 4 seconds</p> <p>Total time = 7 seconds</p> <p><strong>Example 2:</strong></p> <p><strong>Input:</strong> points = [[3,2],[-2,2]]</p> <p><strong>Output:</strong> 5</p> <p><strong>Constraints:</strong></p> <ul> <li><code>points.length == n</code></li> <li><code>1 <= n <= 100</code></li> <li><code>points[i].length == 2</code></li> <li><code>-1000 <= points[i][0], points[i][1] <= 1000</code></li> </ul>
  • Constructor Details

    • Solution

      public Solution()
  • Method Details

    • minTimeToVisitAllPoints

      public int minTimeToVisitAllPoints(int[][] points)