## leetcode 88. Merge Sorted Array（python）

### 描述

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.

Note:

``````nums1.length == m + n
nums2.length == n
0 <= m, n <= 200
1 <= m + n <= 200
-10^9 <= nums1[i], nums2[j] <= 10^9

### 解析

• 当 n 为 0 的时候，表示 nums1 不需要和 nums2 进行合并，不管 nums1 是啥，直接返回 nums1 就可以了

• 当 m 为 0 的时候，表示 nums1 为空，没有需要合并的元素，但是此时不能直接返回 nums2 ，因为题目要求最后返回的是 nums1 ，所以不管 nums2 有多少元素，直接赋值给 nums1 ，然后再返回 nums1 才可以

• 从 nums1 的最后一个有效元素和 nums2 的最后一个有效元素开始逆向比较大小：

• 如果 nums1[m-1] 大于等于 nums2[n-1] ，则将 nums1[m-1] 赋值给 nums1[m+n-1] ，同时 m 减一
• 否则将 nums2[n-1] 赋值给 nums1[m+n-1] ，同时 n 减一
• 如果 nums2 先遍历结束，因为 nums1 剩下的元素一开始就是有序的，所以不用管他，但如果 nums 1 先遍历结束，可能 nums2 还有剩下的元素没有遍历，因为 nums2 剩下的元素一开始就是有序的，直接将 nums2[:n] 赋值给 nums1[:n] 即可

• 返回 nums1

### 解答

``````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.
"""
if n == 0:
return nums1
if m == 0:
nums1[:n] = nums2[:n]
return nums1
while m>0 and n>0:
if nums1[m-1]>=nums2[n-1]:
nums1[m+n-1] = nums1[m-1]
m -= 1
else:
nums1[m+n-1] = nums2[n-1]
n -= 1
nums1[:n] = nums2[:n]
return nums1

### 运行结果

``````Runtime: 38 ms, faster than 20.56% of Python online submissions for Merge Sorted Array.
Memory Usage: 13.6 MB, less than 16.74% of Python online submissions for Merge Sorted Array.