Given two ordered arrays of integers and nums2 nums1, incorporated into the nums2 nums1 in an ordered array such as a num1.
Description:
And initializing the number of elements nums1 nums2 of m and n.
You can assume nums1 sufficient space (space equal to or greater than m + n) to save the elements of nums2.
method one:
Double pointer method, traversing two arrays, one by one from front to back comparison add, require auxiliary space;
The time complexity of O (m + n), the spatial complexity of O (m)
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.
"""
copy_num1 = nums1[:m]
nums1[:] = []
# 双指针
p1 = 0
p2 = 0
while p1 < m and p2 < n:
if copy_num1[p1] < nums2[p2]:
nums1.append(copy_num1[p1])
p1 += 1
else:
nums1.append(nums2[p2])
p2 += 1
if p1 < m:
nums1[p1 + p2:] = copy_num1[p1:]
if p2 < n:
nums1[p1 + p2:] = nums2[p2:]
a = Solution()
num1 = [ 1,2,3,0,0,0]
m=3
n=3
num2=[2,5,6]
a.merge(num1,m,num2,n)
print(num1)Method Two:
In the method a more optimized by topic shows that we can from the back to add from large to small in num1 array. Without auxiliary space.
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.
"""
# 双指针
p1 = m - 1
p2 = n - 1
k = m + n - 1
while p1 >= 0 and p2 >= 0:
if nums1[p1] < nums2[p2]:
nums1[k] = nums2[p2]
p2 -= 1
else:
nums1[k] = nums1[p1]
p1 -= 1
k -= 1
# m<n，加入num2剩余元素
nums1[:p2+1] = nums2[:p2+1]
a = Solution()
num1 = [ 1,2,3,0,0,0]
m=3
n=3
num2=[2,5,6]
a.merge(num1,m,num2,n)
print(num1)
