88. Merge Sorted Array
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Array , Two Pointers , Sorting
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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.
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. """ i = len(nums1) - n for j in nums2: nums1[i] = j i += 1 nums1.sort()