Leetcode|Merge Sorted Array(2つの秩序配列を結合)

Description:
Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array.
The number of elements initialized in nums1 and nums2 are m and n respectively. You may assume that nums1 has a size equal to m + n such that it has enough space to hold additional elements from nums2.
Example:
Input: nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3 Output: [1,2,2,3,5,6]
Input: nums1 = [1], m = 1, nums2 = [], n = 0 Output: [1]
Ideas:

	Oh, first of all, I must remind you guys, this problem return nums1 default. So you cant return user-defined list.
	In python, you should remember sorted() function. Of course we can merge two array and reorder it.
	Second method is using pointer, but after I saw the official one, I would say that most of us havent 
make use of the problem condition. The two array are already sorted and nums1 have n nodes extra space. 
If you traverse the list from back and you can modify nums1 direct, In this way, you can save O(m-1) space complexity.

My Code:

class Solution(object):
    def merge(self, nums1, m, nums2, n):
        """
        :type nums1: List[int]
        :type m: int
        :type nums2: List[int]
        :type n: int
        :rtype: None Do not return anything, modify nums1 in-place instead.
        """
        idx1, idx2 = 0, 0
        nums1_copy = nums1[:m]
        nums1[:] = []
        while idx1 < m and idx2 < n:
            x = nums1_copy[idx1]
            y = nums2[idx2]
            if x > y:
                nums1.append(y)
                idx2 += 1
            else:
                nums1.append(x)
                idx1 += 1

        if idx1 < m:
            nums1.extend(nums1_copy[idx1:])
        elif idx2 < n:
            nums1.extend(nums2[idx2:])

Execution time:28 ms Memory consumption:13.1 MB
Official Method:

def merge(self, nums1, m, nums2, n):
	"""
	:type nums1: List[int]
	:type m: int
	:type nums2: List[int]
	:type n: int
	:rtype: void Do not return anything, modify nums1 in-place instead.
	"""
	# two get pointers for nums1 and nums2
	p1 = m - 1
	p2 = n - 1
	# set pointer for nums1
	p = m + n - 1
	
	# while there are still elements to compare
	while p1 >= 0 and p2 >= 0:
	    if nums1[p1] < nums2[p2]:
	        nums1[p] = nums2[p2]
	        p2 -= 1
	    else:
	        nums1[p] =  nums1[p1]
	        p1 -= 1
	    p -= 1
	
	# add missing elements from nums2
	nums1[:p2 + 1] = nums2[:p2 + 1]

  ：LeetCode
  ：https://leetcode-cn.com/problems/merge-sorted-array/solution/he-bing-liang-ge-you-xu-shu-zu-by-leetcode/
  ：  （LeetCode）

Execution time:20 ms Memery consumption:13.1 MB
summary

	 If we can modify it in origional sapce, try not to create extra space to solve problem.