You are given two integer arrays nums1 and nums2, sorted in non-decreasing order, and two integers m and n, representing the number of elements in nums1 and nums2 respectively.
Merge nums1 and nums2 into a single array sorted in non-decreasing order.
The final sorted array should not be returned by the function, but instead be stored inside the array nums1. To accommodate this, nums1 has a length of m + n, where the first m elements denote the elements that should be merged, and the last n elements are set to 0 and should be ignored. nums2 has a length of n.
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Example 1:
\n\n\nInput: nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3\nOutput: [1,2,2,3,5,6]\nExplanation: The arrays we are merging are [1,2,3] and [2,5,6].\nThe result of the merge is [1,2,2,3,5,6] with the underlined elements coming from nums1.\n\n\n
Example 2:
\n\n\nInput: nums1 = [1], m = 1, nums2 = [], n = 0\nOutput: [1]\nExplanation: The arrays we are merging are [1] and [].\nThe result of the merge is [1].\n\n\n
Example 3:
\n\n\nInput: nums1 = [0], m = 0, nums2 = [1], n = 1\nOutput: [1]\nExplanation: 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\n\n
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Constraints:
\n\nnums1.length == m + nnums2.length == n0 <= m, n <= 2001 <= m + n <= 200-109 <= nums1[i], nums2[j] <= 109\n
Follow up: Can you come up with an algorithm that runs in O(m + n) time?
给你两个按 非递减顺序 排列的整数数组 nums1 和 nums2，另有两个整数 m 和 n ，分别表示 nums1 和 nums2 中的元素数目。
请你 合并 nums2 到 nums1 中，使合并后的数组同样按 非递减顺序 排列。
注意：最终，合并后数组不应由函数返回，而是存储在数组 nums1 中。为了应对这种情况，nums1 的初始长度为 m + n，其中前 m 个元素表示应合并的元素，后 n 个元素为 0 ，应忽略。nums2 的长度为 n 。
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示例 1：
\n\n\n输入：nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3\n输出：[1,2,2,3,5,6]\n解释：需要合并 [1,2,3] 和 [2,5,6] 。\n合并结果是 [1,2,2,3,5,6] ，其中斜体加粗标注的为 nums1 中的元素。\n\n\n
示例 2：
\n\n\n输入：nums1 = [1], m = 1, nums2 = [], n = 0\n输出：[1]\n解释：需要合并 [1] 和 [] 。\n合并结果是 [1] 。\n\n\n
示例 3：
\n\n\n输入：nums1 = [0], m = 0, nums2 = [1], n = 1\n输出：[1]\n解释：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\n\n
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提示：
\n\nnums1.length == m + nnums2.length == n0 <= m, n <= 2001 <= m + n <= 200-109 <= nums1[i], nums2[j] <= 109\n\n
进阶：你可以设计实现一个时间复杂度为 O(m + n) 的算法解决此问题吗？
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