leetcode-problemset/leetcode-cn/originData/merge-sorted-array.json

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    "data": {
        "question": {
            "questionId": "88",
            "questionFrontendId": "88",
            "categoryTitle": "Algorithms",
            "boundTopicId": 1158,
            "title": "Merge Sorted Array",
            "titleSlug": "merge-sorted-array",
            "content": "<p>You are given two integer arrays <code>nums1</code> and <code>nums2</code>, sorted in <strong>non-decreasing order</strong>, and two integers <code>m</code> and <code>n</code>, representing the number of elements in <code>nums1</code> and <code>nums2</code> respectively.</p>\n\n<p><strong>Merge</strong> <code>nums1</code> and <code>nums2</code> into a single array sorted in <strong>non-decreasing order</strong>.</p>\n\n<p>The final sorted array should not be returned by the function, but instead be <em>stored inside the array </em><code>nums1</code>. To accommodate this, <code>nums1</code> has a length of <code>m + n</code>, where the first <code>m</code> elements denote the elements that should be merged, and the last <code>n</code> elements are set to <code>0</code> and should be ignored. <code>nums2</code> has a length of <code>n</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3\n<strong>Output:</strong> [1,2,2,3,5,6]\n<strong>Explanation:</strong> The arrays we are merging are [1,2,3] and [2,5,6].\nThe result of the merge is [<u>1</u>,<u>2</u>,2,<u>3</u>,5,6] with the underlined elements coming from nums1.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums1 = [1], m = 1, nums2 = [], n = 0\n<strong>Output:</strong> [1]\n<strong>Explanation:</strong> The arrays we are merging are [1] and [].\nThe result of the merge is [1].\n</pre>\n\n<p><strong>Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums1 = [0], m = 0, nums2 = [1], n = 1\n<strong>Output:</strong> [1]\n<strong>Explanation:</strong> The arrays we are merging are [] and [1].\nThe result of the merge is [1].\nNote that because m = 0, there are no elements in nums1. The 0 is only there to ensure the merge result can fit in nums1.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>nums1.length == m + n</code></li>\n\t<li><code>nums2.length == n</code></li>\n\t<li><code>0 &lt;= m, n &lt;= 200</code></li>\n\t<li><code>1 &lt;= m + n &lt;= 200</code></li>\n\t<li><code>-10<sup>9</sup> &lt;= nums1[i], nums2[j] &lt;= 10<sup>9</sup></code></li>\n</ul>\n\n<p>&nbsp;</p>\n<p><strong>Follow up: </strong>Can you come up with an algorithm that runs in <code>O(m + n)</code> time?</p>\n",
            "translatedTitle": "合并两个有序数组",
            "translatedContent": "<p>给你两个按 <strong>非递减顺序</strong> 排列的整数数组&nbsp;<code>nums1</code><em> </em>和 <code>nums2</code>，另有两个整数 <code>m</code> 和 <code>n</code> ，分别表示 <code>nums1</code> 和 <code>nums2</code> 中的元素数目。</p>\n\n<p>请你 <strong>合并</strong> <code>nums2</code><em> </em>到 <code>nums1</code> 中，使合并后的数组同样按 <strong>非递减顺序</strong> 排列。</p>\n\n<p><strong>注意：</strong>最终，合并后数组不应由函数返回，而是存储在数组 <code>nums1</code> 中。为了应对这种情况，<code>nums1</code> 的初始长度为 <code>m + n</code>，其中前 <code>m</code> 个元素表示应合并的元素，后 <code>n</code> 个元素为 <code>0</code> ，应忽略。<code>nums2</code> 的长度为 <code>n</code> 。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3\n<strong>输出：</strong>[1,2,2,3,5,6]\n<strong>解释：</strong>需要合并 [1,2,3] 和 [2,5,6] 。\n合并结果是 [<em><strong>1</strong></em>,<em><strong>2</strong></em>,2,<em><strong>3</strong></em>,5,6] ，其中斜体加粗标注的为 nums1 中的元素。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums1 = [1], m = 1, nums2 = [], n = 0\n<strong>输出：</strong>[1]\n<strong>解释：</strong>需要合并 [1] 和 [] 。\n合并结果是 [1] 。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums1 = [0], m = 0, nums2 = [1], n = 1\n<strong>输出：</strong>[1]\n<strong>解释：</strong>需要合并的数组是 [] 和 [1] 。\n合并结果是 [1] 。\n注意，因为 m = 0 ，所以 nums1 中没有元素。nums1 中仅存的 0 仅仅是为了确保合并结果可以顺利存放到 nums1 中。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>nums1.length == m + n</code></li>\n\t<li><code>nums2.length == n</code></li>\n\t<li><code>0 &lt;= m, n &lt;= 200</code></li>\n\t<li><code>1 &lt;= m + n &lt;= 200</code></li>\n\t<li><code>-10<sup>9</sup> &lt;= nums1[i], nums2[j] &lt;= 10<sup>9</sup></code></li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>进阶：</strong>你可以设计实现一个时间复杂度为 <code>O(m + n)</code> 的算法解决此问题吗？</p>\n",
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                    "code": "class Solution(object):\n    def merge(self, nums1, m, nums2, n):\n        \"\"\"\n        :type nums1: List[int]\n        :type m: int\n        :type nums2: List[int]\n        :type n: int\n        :rtype: None Do not return anything, modify nums1 in-place instead.\n        \"\"\"",
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                "You can easily solve this problem if you simply think about two elements at a time rather than two arrays. We know that each of the individual arrays is sorted. What we don't know is how they will intertwine. Can we take a local decision and arrive at an optimal solution?",
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