動機
從背後來就是easy，從正面來就是地獄
Problem
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 = 3Output: [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 = 0Output: [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 = 1Output: [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 + nnums2.length == n0 <= m, n <= 2001 <= 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?
Sol
class Solution:
def merge(self, ns1: List[int], m: int, ns2: List[int], n: int) -> None:
i, a, b = m+n-1, m-1, n-1
while i >= 0:
if b >= 0 and a >= 0:
if ns2[b] > ns1[a]:
ns1[i], i = ns2[b], i-1
b -= 1
else:
ns1[i], i = ns1[a], i-1
a -= 1
elif b >= 0:
ns1[i], i = ns2[b], i-1
b -= 1
else:
ns1[i], i = ns1[a], i-1
a -= 1