You are given two integer arrays
nums2, sorted in non-decreasing order, and two integers
n, representing the number of elements in
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
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] with the underlined elements coming from nums1.
Input: nums1 = , m = 1, nums2 = , n = 0 Output:  Explanation: The arrays we are merging are  and . The result of the merge is .
Input: nums1 = , m = 0, nums2 = , n = 1 Output:  Explanation: The arrays we are merging are  and . The result of the merge is . 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.
nums1.length == m + n
nums2.length == n
0 <= m, n <= 200
1 <= m + n <= 200
-109 <= nums1[i], nums2[j] <= 109
Follow up: Can you come up with an algorithm that runs in
O(m + n) time?
Create a copy of nums1 to keep the original values. Then set one pointer at the beginning of nums1, another pointer at the beginning of nums2, and push the smallest value in the nums1 array at each step.
class Solution: def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None: """ Do not return anything, modify nums1 in-place instead. """ nums1_copy = list(nums1) i = 0 j = 0 index = 0 while i < m and j < n: if nums1_copy[i] < nums2[j]: nums1[index] = nums1_copy[i] i += 1 else: nums1[index] = nums2[j] j += 1 index += 1 while i < m: nums1[index] = nums1_copy[i] i += 1 index += 1 while j < n: nums1[index] = nums2[j] j += 1 index += 1
- Time complexity: O(m + n).
- Space complexity: O(m + n).