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88.merge-sorted-array.py
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# Tag: Array, Two Pointers, Sorting
# Time: O(M + N)
# Space: O(1)
# Ref: -
# Note: -
# 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?
#
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.
"""
i = m - 1
j = n - 1
k = m + n - 1
while i >=0 and j >= 0:
if nums1[i] >= nums2[j]:
nums1[k] = nums1[i]
i -= 1
else:
nums1[k] = nums2[j]
j -= 1
k -= 1
while j >= 0:
nums1[k] = nums2[j]
j -= 1
k -= 1
# because if nums1 left, it's already in place