{
  "type": "code_options",
  "author": "https://github.com/qiyuangong/leetcode",
  "source": "solution.md",
  "exercise_id": "6ac5ec62fe274644980c8fbc4bcb5612",
  "keywords": "数组,双指针,排序",
  "title": "合并两个有序数组",
  "desc": [
    {
      "content": "\n<p>给你两个有序整数数组 <code>nums1</code><em> </em>和 <code>nums2</code>，请你将 <code>nums2</code><em> </em>合并到 <code>nums1</code><em> </em>中<em>，</em>使 <code>nums1</code><em> </em>成为一个有序数组。</p><p>初始化 <code>nums1</code> 和 <code>nums2</code> 的元素数量分别为 <code>m</code> 和 <code>n</code><em> </em>。你可以假设 <code>nums1</code><em> </em>的空间大小等于 <code>m + n</code>，这样它就有足够的空间保存来自 <code>nums2</code> 的元素。</p><p> </p><p><strong>示例 1：</strong></p><pre><strong>输入：</strong>nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3<strong><br />输出：</strong>[1,2,2,3,5,6]</pre><p><strong>示例 2：</strong></p><pre><strong>输入：</strong>nums1 = [1], m = 1, nums2 = [], n = 0<strong><br />输出：</strong>[1]</pre><p> </p><p><strong>提示：</strong></p><ul>\t<li><code>nums1.length == m + n</code></li>\t<li><code>nums2.length == n</code></li>\t<li><code>0 <= m, n <= 200</code></li>\t<li><code>1 <= m + n <= 200</code></li>\t<li><code>-10<sup>9</sup> <= nums1[i], nums2[i] <= 10<sup>9</sup></code></li></ul>",
      "language": "markdown"
    }
  ],
  "answer": [
    {
      "content": "",
      "language": "python"
    }
  ],
  "prepared": [
    [
      {
        "content": "",
        "language": "python"
      }
    ],
    [
      {
        "content": "",
        "language": "python"
      }
    ],
    [
      {
        "content": "",
        "language": "python"
      }
    ]
  ],
  "template": {
    "content": "class Solution(object):\n\tdef merge(self, nums1, m, nums2, n):\n\t\t\"\"\"\n\t\t:type nums1: List[int]\n\t\t:type m: int\n\t\t:type nums2: List[int]\n\t\t:type n: int\n\t\t:rtype: void Do not return anything, modify nums1 in-place instead.\n\t\t\"\"\"\n\t\tp1, p2 = m - 1, n - 1\n\t\tpos = m + n - 1\n\t\twhile p1 >= 0 and p2 >= 0:\n\t\t\tif nums1[p1] >= nums2[p2]:\n\t\t\t\tnums1[pos] = nums1[p1]\n\t\t\t\tp1 -= 1\n\t\t\telse:\n\t\t\t\tnums1[pos] = nums2[p2]\n\t\t\t\tp2 -= 1\n\t\t\tpos -= 1\n\t\twhile p2 >= 0:\n\t\t\tnums1[pos] = nums2[p2]\n\t\t\tp2 -= 1\n\t\t\tpos -= 1\n\t\treturn nums1\n# %%\ns = Solution()\nprint(s.merge(nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3))",
    "language": "python"
  },
  "node_id": "dailycode-8c71087921b345cd8aceed9913f3cea3",
  "license": "csdn.net",
  "created_at": 1637894161,
  "topic_link": "https://bbs.csdn.net/topics/600470926"
}