Merge Sorted Array

88. Merge Sorted Array

Difficulty: Easy

Related Topics: Array, Two Pointers, Sorting

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] with the underlined elements coming from nums1.

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]. 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.

Code:

class Solution:
    def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
        if n == 0: return nums1
        for i in range(n):
            nums1[-(i+1)] = nums2[i]
        nums1.sort()
        return nums1

Feedbacks:

I return original list nums1 if nums2 has no element in it. Else, I looped for n times to replace 0s to element of nums2 because there are same n numbers of 0s in nums1 and numbers to be merged in nums2. And then I sorted the list before I return the nums1 which is the answer. Runtime is quite okay, but memory occupation could be improved.