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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.
Example 1:
<pre>Input: nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3 Output: [1,2,2,3,5,6] Explanation: The arrays we are merging are [1,2,3] and [2,5,6]. The result of the merge is [<u>1</u>,<u>2</u>,2,<u>3</u>,5,6] with the underlined elements coming from nums1. </pre>
Example 2:
<pre>Input: nums1 = [1], m = 1, nums2 = [], n = 0 Output: [1] Explanation: The arrays we are merging are [1] and []. The result of the merge is [1]. </pre>
Example 3:
<pre>Input: nums1 = [0], m = 0, nums2 = [1], n = 1 Output: [1] Explanation: The arrays we are merging are [] and [1]. The result of the merge is [1]. Note that because m = 0, there are no elements in nums1. The 0 is only there to ensure the merge result can fit in nums1. </pre>
Constraints:
nums1.length == m + nnums2.length == n0 <= m, n <= 2001 <= m + n <= 200-10<sup>9</sup> <= nums1[i], nums2[j] <= 10<sup>9</sup>
Follow up: Can you come up with an algorithm that runs in O(m + n) time?
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题目大意：
合并两有序数组，最后结果储存在第一个数组
解题思路：
从后往前合并
解题步骤：
N/A
注意事项：
- i从m - 1而不是len(nums1) - 1开始，m和n是数组实际长度。‘
- i小于0就取j那一边，而nums1[i] < nums2[j]就必须保证j也大于等于0，i肯定是了。
算法I解题思路(推荐)：
Python代码：
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14def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
"""
Do not return anything, modify nums1 in-place instead.
"""
i, j, k= m - 1, n - 1, len(nums1) - 1
while i >= 0 or j >= 0:
if i < 0 or (j >= 0 and nums1[i] < nums2[j]):
nums1[k] = nums2[j]
j -= 1
k -= 1
else:
nums1[k] = nums1[i]
i -= 1
k -= 1
算法II解题思路：
Python代码：
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19def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
i, j, k = m - 1, n - 1, len(nums1) - 1
while i >= 0 and j >= 0:
if nums1[i] > nums2[j]:
nums1[k] = nums1[i]
k -= 1
i -= 1
else:
nums1[k] = nums2[j]
k -= 1
j -= 1
while i >= 0:
nums1[k] = nums1[i]
k -= 1
i -= 1
while j >= 0:
nums1[k] = nums2[j]
k -= 1
j -= 1
算法分析：
时间复杂度为O(n + m)，空间复杂度O(1)
