leetcode-problemset/leetcode-cn/originData/keep-multiplying-found-values-by-two.json

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            "questionId": "2274",
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            "categoryTitle": "Algorithms",
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            "title": "Keep Multiplying Found Values by Two",
            "titleSlug": "keep-multiplying-found-values-by-two",
            "content": "<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>\n\n<p>You then do the following steps:</p>\n\n<ol>\n\t<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>\n\t<li>Otherwise, <strong>stop</strong> the process.</li>\n\t<li><strong>Repeat</strong> this process with the new number as long as you keep finding the number.</li>\n</ol>\n\n<p>Return <em>the <strong>final</strong> value of </em><code>original</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [5,3,6,1,12], original = 3\n<strong>Output:</strong> 24\n<strong>Explanation:</strong> \n- 3 is found in nums. 3 is multiplied by 2 to obtain 6.\n- 6 is found in nums. 6 is multiplied by 2 to obtain 12.\n- 12 is found in nums. 12 is multiplied by 2 to obtain 24.\n- 24 is not found in nums. Thus, 24 is returned.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [2,7,9], original = 4\n<strong>Output:</strong> 4\n<strong>Explanation:</strong>\n- 4 is not found in nums. Thus, 4 is returned.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 1000</code></li>\n\t<li><code>1 &lt;= nums[i], original &lt;= 1000</code></li>\n</ul>\n",
            "translatedTitle": "将找到的值乘以 2",
            "translatedContent": "<p>给你一个整数数组 <code>nums</code> ，另给你一个整数 <code>original</code> ，这是需要在 <code>nums</code> 中搜索的第一个数字。</p>\n\n<p>接下来，你需要按下述步骤操作：</p>\n\n<ol>\n\t<li>如果在 <code>nums</code> 中找到 <code>original</code> ，将 <code>original</code>&nbsp;<strong>乘以</strong> 2 ，得到新 <code>original</code>（即，令 <code>original = 2 * original</code>）。</li>\n\t<li>否则，停止这一过程。</li>\n\t<li>只要能在数组中找到新 <code>original</code> ，就对新 <code>original</code> 继续 <strong>重复</strong> 这一过程<strong>。</strong></li>\n</ol>\n\n<p>返回<em> </em><code>original</code> 的 <strong>最终</strong> 值。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [5,3,6,1,12], original = 3\n<strong>输出：</strong>24\n<strong>解释：</strong> \n- 3 能在 nums 中找到。3 * 2 = 6 。\n- 6 能在 nums 中找到。6 * 2 = 12 。\n- 12 能在 nums 中找到。12 * 2 = 24 。\n- 24 不能在 nums 中找到。因此，返回 24 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [2,7,9], original = 4\n<strong>输出：</strong>4\n<strong>解释：</strong>\n- 4 不能在 nums 中找到。因此，返回 4 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 1000</code></li>\n\t<li><code>1 &lt;= nums[i], original &lt;= 1000</code></li>\n</ul>\n",
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                "Repeatedly iterate through the array and check if the current value of original is in the array.",
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