LeetCode 088 Merge Sorted Array

LeetCode

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
-10<sup>9</sup> <= nums1[i], nums2[j] <= 10<sup>9</sup>

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

题目大意：

合并两有序数组，最后结果储存在第一个数组

解题思路：

从后往前合并

解题步骤：

N/A

注意事项：

i从m - 1而不是len(nums1) - 1开始，m和n是数组实际长度。

Python代码：

def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
	i, j, k = m - 1, n - 1, len(nums1) - 1
	while i >= 0 and j >= 0:
		if nums1[i] > nums2[j]:
			nums1[k] = nums1[i]
			k -= 1
			i -= 1
		else:
			nums1[k] = nums2[j]
			k -= 1
			j -= 1
	while i >= 0:
		nums1[k] = nums1[i]
		k -= 1
		i -= 1
	while j >= 0:
		nums1[k] = nums2[j]
		k -= 1
		j -= 1

算法分析：

时间复杂度为O(n + m)，空间复杂度O(1)