leetcode-problemset/leetcode-cn/originData/distribute-elements-into-two-arrays-i.json

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            "title": "Distribute Elements Into Two Arrays I",
            "titleSlug": "distribute-elements-into-two-arrays-i",
            "content": "<p>You are given a <strong>1-indexed</strong> array of <strong>distinct</strong> integers <code>nums</code> of length <code>n</code>.</p>\n\n<p>You need to distribute all the elements of <code>nums</code> between two arrays <code>arr1</code> and <code>arr2</code> using <code>n</code> operations. In the first operation, append <code>nums[1]</code> to <code>arr1</code>. In the second operation, append <code>nums[2]</code> to <code>arr2</code>. Afterwards, in the <code>i<sup>th</sup></code> operation:</p>\n\n<ul>\n\t<li>If the last element of <code>arr1</code> is<strong> greater</strong> than the last element of <code>arr2</code>, append <code>nums[i]</code> to <code>arr1</code>. Otherwise, append <code>nums[i]</code> to <code>arr2</code>.</li>\n</ul>\n\n<p>The array <code>result</code> is formed by concatenating the arrays <code>arr1</code> and <code>arr2</code>. For example, if <code>arr1 == [1,2,3]</code> and <code>arr2 == [4,5,6]</code>, then <code>result = [1,2,3,4,5,6]</code>.</p>\n\n<p>Return <em>the array</em> <code>result</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [2,1,3]\n<strong>Output:</strong> [2,3,1]\n<strong>Explanation:</strong> After the first 2 operations, arr1 = [2] and arr2 = [1].\nIn the 3<sup>rd</sup> operation, as the last element of arr1 is greater than the last element of arr2 (2 &gt; 1), append nums[3] to arr1.\nAfter 3 operations, arr1 = [2,3] and arr2 = [1].\nHence, the array result formed by concatenation is [2,3,1].\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [5,4,3,8]\n<strong>Output:</strong> [5,3,4,8]\n<strong>Explanation:</strong> After the first 2 operations, arr1 = [5] and arr2 = [4].\nIn the 3<sup>rd</sup> operation, as the last element of arr1 is greater than the last element of arr2 (5 &gt; 4), append nums[3] to arr1, hence arr1 becomes [5,3].\nIn the 4<sup>th</sup> operation, as the last element of arr2 is greater than the last element of arr1 (4 &gt; 3), append nums[4] to arr2, hence arr2 becomes [4,8].\nAfter 4 operations, arr1 = [5,3] and arr2 = [4,8].\nHence, the array result formed by concatenation is [5,3,4,8].\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>3 &lt;= n &lt;= 50</code></li>\n\t<li><code>1 &lt;= nums[i] &lt;= 100</code></li>\n\t<li>All elements in <code>nums</code> are distinct.</li>\n</ul>\n",
            "translatedTitle": "将元素分配到两个数组中 I",
            "translatedContent": "<p>给你一个下标从 <strong>1</strong> 开始、包含<strong> 不同 </strong>整数的数组 <code>nums</code> ，数组长度为 <code>n</code> 。</p>\n\n<p>你需要通过 <code>n</code> 次操作，将 <code>nums</code> 中的所有元素分配到两个数组 <code>arr1</code> 和 <code>arr2</code> 中。在第一次操作中，将 <code>nums[1]</code> 追加到 <code>arr1</code> 。在第二次操作中，将 <code>nums[2]</code> 追加到 <code>arr2</code> 。之后，在第 <code>i</code> 次操作中：</p>\n\n<ul>\n\t<li>如果 <code>arr1</code> 的最后一个元素 <strong>大于 </strong><code>arr2</code> 的最后一个元素，就将 <code>nums[i]</code> 追加到 <code>arr1</code> 。否则，将 <code>nums[i]</code> 追加到 <code>arr2</code> 。</li>\n</ul>\n\n<p>通过连接数组 <code>arr1</code> 和 <code>arr2</code> 形成数组 <code>result</code> 。例如，如果 <code>arr1 == [1,2,3]</code> 且 <code>arr2 == [4,5,6]</code> ，那么 <code>result = [1,2,3,4,5,6]</code> 。</p>\n\n<p>返回数组 <code>result</code> 。</p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [2,1,3]\n<strong>输出：</strong>[2,3,1]\n<strong>解释：</strong>在前两次操作后，arr1 = [2] ，arr2 = [1] 。\n在第 3 次操作中，由于 arr1 的最后一个元素大于 arr2 的最后一个元素（2 &gt; 1），将 nums[3] 追加到 arr1 。\n3 次操作后，arr1 = [2,3] ，arr2 = [1] 。\n因此，连接形成的数组 result 是 [2,3,1] 。\n</pre>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [5,4,3,8]\n<strong>输出：</strong>[5,3,4,8]\n<strong>解释：</strong>在前两次操作后，arr1 = [5] ，arr2 = [4] 。\n在第 3 次操作中，由于 arr1 的最后一个元素大于 arr2 的最后一个元素（5 &gt; 4），将 nums[3] 追加到 arr1 ，因此 arr1 变为 [5,3] 。\n在第 4 次操作中，由于 arr2 的最后一个元素大于 arr1 的最后一个元素（4 &gt; 3），将 nums[4] 追加到 arr2 ，因此 arr2 变为 [4,8] 。\n4 次操作后，arr1 = [5,3] ，arr2 = [4,8] 。\n因此，连接形成的数组 result 是 [5,3,4,8] 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>3 &lt;= n &lt;= 50</code></li>\n\t<li><code>1 &lt;= nums[i] &lt;= 100</code></li>\n\t<li><code>nums</code>中的所有元素都互不相同。</li>\n</ul>\n",
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