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?
The size of nums1 is m + n. nums1 is enough to store all numbers in nums1 and nums2. We only need to first shift all numbers in nums1 by n positions. Then we compare and put numbers in nums1 and nums2 into nums1 in ascending order.
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. """ for i in range(m - 1, -1, -1): nums1[i + n] = nums1[i] i = n j = 0 k = 0 while i < m + n and j < n: if nums1[i] < nums2[j]: nums1[k] = nums1[i] i += 1 else: nums1[k] = nums2[j] j += 1 k += 1 while j < n: nums1[k] = nums2[j] j += 1 k += 1
- Time complexity: O(m + n).
- Space complexity: O(m + n).