- Merge Sorted Array 难度 简单
https://leetcode-cn.com/problems/merge-sorted-array/
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
Follow up: Can you come up with an algorithm that runs in O(m + n) time?
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相关标签
- Array
- Two Pointers
- Sorting
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隐藏提示1 You can easily solve this problem if you simply think about two elements at a time rather than two arrays. We know that each of the individual arrays is sorted. What we don't know is how they will intertwine. Can we take a local decision and arrive at an optimal solution?
显示提示2 If you simply consider one element each at a time from the two arrays and make a decision and proceed accordingly, you will arrive at the optimal solution.
# time O(n)
class Solution:
def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
"""
Do not return anything, modify nums1 in-place instead.
"""
n1copy = list(nums1[:m])
n1copy_i = 0
n2_i = 0
n1_i = 0
while n1copy_i < m and n2_i < n and n1_i < m+n:
if n1copy[n1copy_i] <= nums2[n2_i]:
nums1[n1_i] = n1copy[n1copy_i]
n1copy_i += 1
else:
nums1[n1_i] = nums2[n2_i]
n2_i += 1
n1_i += 1
while n1copy_i >= m and n2_i < n and n1_i < m+n:
nums1[n1_i] = nums2[n2_i]
n2_i += 1
n1_i += 1
while n2_i >= n and n1copy_i < m and n1_i < m+n:
nums1[n1_i] = n1copy[n1copy_i]
n1copy_i += 1
n1_i += 1# time O(n)
class Solution:
def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
"""
Do not return anything, modify nums1 in-place instead.
"""
if nums1 is None or nums2 is None:
return
i, j = 0, 0
array = []
while i < m and j < n:
if nums1[i] <= nums2[j]:
array.append(nums1[i])
i += 1
else:
array.append(nums2[j])
j += 1
while i < m:
array.append(nums1[i])
i += 1
while j < n:
array.append(nums2[j])
j += 1
for index in range(m + n):
nums1[index] = array[index]
return C = A+B
C.sort()
return C