[LeetCode] 88. Merge Sorted Array (Python)

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88. Merge Sorted Array

Easy

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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.

 

Constraints:

• nums1.length == m + n
• nums2.length == n
• 0 <= m, n <= 200
• 1 <= m + n <= 200
• -109 <= nums1[i], nums2[j] <= 109

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Code

class Solution(object):
    def merge(self, nums1, m, nums2, n):
        for i in range(m, m+n):
            nums1[i] = nums2[i-m]
        nums1.sort()        

Time: 0 ms (100%), Space: 12.5 MB (13.5%)

Follow Up

Follow up: Can you come up with an algorithm that runs in O(m + n) time?

• Loop: $O(n)$
• Sorting: $O((m+n) log(m + n))$

Time Complexity

$O((m+n) log(m + n))$

내가 작성한 코드의 시간 복잡도는 $O((m+n) log(m + n))$이다.

그렇다면, 문제에서 요구하는 시간 복잡도 $O(m+n)$로 푸는 방법은 무엇이 있을까?

Two-Pointer Solution

투 포인터 방법으로 $O(m+n)$ 시간 복잡도를 달성할 수 있다.

class Solution(object):
    def merge(self, nums1, m, nums2, n):
        p1 = m - 1 
        p2 = n - 1 
        p = m + n - 1  

        while p1 >= 0 and p2 >= 0:
            if nums1[p1] > nums2[p2]:
                nums1[p] = nums1[p1]
                p1 -= 1
            else:
                nums1[p] = nums2[p2]
                p2 -= 1
            p -= 1
        
        while p2 >= 0:
            nums1[p] = nums2[p2]
            p2 -= 1
            p -= 1