(https://leetcode.com/problems/merge-sorted-array/)
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:
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 [1,2,2,3,5,6].
Example 2:
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].
Example 3:
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].
Constraints:
- nums1.length == m + n
- nums2.length == n
- 0 <= m, n <= 200
- 1 <= m + n <= 200
- -10⁹ <= nums1[i], nums2[j] <= 10⁹
- array
- two pointers
def merge(nums1, m, nums2, n):
# fill from the end
i, j, k = m - 1, n - 1, m + n - 1
while j >= 0:
if i >= 0 and nums1[i] > nums2[j]:
nums1[k] = nums1[i]
i -= 1
else:
nums1[k] = nums2[j]
j -= 1
k -= 1Time complexity : O(m + n)
Space complexity : O(1)