robot-return-to-origin

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Algorithms

There is a robot starting at the position <code>(0, 0)</code>, the origin, on a 2D plane. Given a sequence of its moves, judge if this robot ends up at <code>(0, 0)</code> after it completes its moves.

You are given a string <code>moves</code> that represents the move sequence of the robot where <code>moves[i]</code> represents its <code>ith</code> move. Valid moves are <code>&#39;R&#39;</code> (right), <code>&#39;L&#39;</code> (left), <code>&#39;U&#39;</code> (up), and <code>&#39;D&#39;</code> (down).

Return <code>true</code> if the robot returns to the origin after it finishes all of its moves, or <code>false</code> otherwise.

Note: The way that the robot is &quot;facing&quot; is irrelevant. <code>&#39;R&#39;</code> will always make the robot move to the right once, <code>&#39;L&#39;</code> will always make it move left, etc. Also, assume that the magnitude of the robot&#39;s movement is the same for each move.

&nbsp;
Example 1:

<pre>
Input: moves = &quot;UD&quot;
Output: true
Explanation: The robot moves up once, and then down once. All moves have the same magnitude, so it ended up at the origin where it started. Therefore, we return true.
</pre>

Example 2:

<pre>
Input: moves = &quot;LL&quot;
Output: false
Explanation: The robot moves left twice. It ends up two &quot;moves&quot; to the left of the origin. We return false because it is not at the origin at the end of its moves.
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= moves.length &lt;= 2 * 104</code></li>
	<li><code>moves</code> only contains the characters <code>&#39;U&#39;</code>, <code>&#39;D&#39;</code>, <code>&#39;L&#39;</code> and <code>&#39;R&#39;</code>.</li>
</ul>

judgeCircle

Robot Return to Origin

Number of Provinces

number-of-provinces

省份数量

Execution of All Suffix Instructions Staying in a Grid

execution-of-all-suffix-instructions-staying-in-a-grid

执行所有后缀指令

Furthest Point From Origin

furthest-point-from-origin

距离原点最远的点

String

Simulation

在二维平面上，有一个机器人从原点 <code>(0, 0)</code> 开始。给出它的移动顺序，判断这个机器人在完成移动后是否在&nbsp;<code>(0, 0)</code> 处结束。

移动顺序由字符串&nbsp;<code>moves</code>&nbsp;表示。字符 <code>move[i]</code> 表示其第 <code>i</code> 次移动。机器人的有效动作有&nbsp;<code>R</code>（右），<code>L</code>（左），<code>U</code>（上）和 <code>D</code>（下）。

如果机器人在完成所有动作后返回原点，则返回 <code>true</code>。否则，返回 <code>false</code>。

注意：机器人“面朝”的方向无关紧要。 <code>“R”</code> 将始终使机器人向右移动一次，<code>“L”</code> 将始终向左移动等。此外，假设每次移动机器人的移动幅度相同。

&nbsp;

示例 1:

<pre>
输入: moves = "UD"
输出: true
解释：机器人向上移动一次，然后向下移动一次。所有动作都具有相同的幅度，因此它最终回到它开始的原点。因此，我们返回 true。</pre>

示例 2:

<pre>
输入: moves = "LL"
输出: false
解释：机器人向左移动两次。它最终位于原点的左侧，距原点有两次 “移动” 的距离。我们返回 false，因为它在移动结束时没有返回原点。</pre>

&nbsp;

提示:

<ul>
	<li><code>1 &lt;= moves.length &lt;= 2 * 104</code></li>
	<li><code>moves</code>&nbsp;只包含字符&nbsp;<code>'U'</code>,&nbsp;<code>'D'</code>,&nbsp;<code>'L'</code>&nbsp;和&nbsp;<code>'R'</code></li>
</ul>