You are given two integer arrays nums1
and nums2
, sorte 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 sorted 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
.
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 mergin 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.
Input : nums1 = [1], m = 1, nums2 = [], n = 0
Output : [1]
Explanation : The arrays we are mergin are [1] and [], The result of the merge is [1].
Input : nums1 = [0], m = 0, nums2 = [1], n = 1
Output : [1]
Explanation : The arrays we are mergin 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.
nums1.length == m+n
nums2.length == n
nums1 = nums[:m+1] (x : 새로운 참조 할당)
del nums1[m:] (o : nums1 자체에서 값만 제거)
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.
"""
del nums1[m:] # nums1 길이 m만 남겨둔다.
nums1 += nums2[0:n] # nums2 길이는 n
nums1.sort()