国外版

算法题（国外版）/keep-multiplying-found-values-by-two.html

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+<p>You are given an array of integers <code>nums</code>. You are also given an integer <code>original</code> which is the first number that needs to be searched for in <code>nums</code>.</p>
+
+<p>You then do the following steps:</p>
+
+<ol>
+	<li>If <code>original</code> is found in <code>nums</code>, <strong>multiply</strong> it by two (i.e., set <code>original = 2 * original</code>).</li>
+	<li>Otherwise, <strong>stop</strong> the process.</li>
+	<li><strong>Repeat</strong> this process with the new number as long as you keep finding the number.</li>
+</ol>
+
+<p>Return <em>the <strong>final</strong> value of </em><code>original</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [5,3,6,1,12], original = 3
+<strong>Output:</strong> 24
+<strong>Explanation:</strong> 
+- 3 is found in nums. 3 is multiplied by 2 to obtain 6.
+- 6 is found in nums. 6 is multiplied by 2 to obtain 12.
+- 12 is found in nums. 12 is multiplied by 2 to obtain 24.
+- 24 is not found in nums. Thus, 24 is returned.
+</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [2,7,9], original = 4
+<strong>Output:</strong> 4
+<strong>Explanation:</strong>
+- 4 is not found in nums. Thus, 4 is returned.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
+	<li><code>1 &lt;= nums[i], original &lt;= 1000</code></li>
+</ul>