Class Solution
java.lang.Object
g1201_1300.s1266_minimum_time_visiting_all_points.Solution
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>
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Constructor Summary
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Constructor Details
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Solution
public Solution()
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Method Details
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minTimeToVisitAllPoints
public int minTimeToVisitAllPoints(int[][] points)
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