以下解决方案的时间复杂度是 O(N) 吗?
is the following solutions' time complexity O(N)?
我试着计算函数的时间复杂度"twoSum"并写下每行的成本和它执行的次数。最后,我制作了两个向量:一个用于成本,另一个用于频率,然后计算它们的内积。我得到以下信息:(n + n + nlogn + n + n + nlogn + n + n + n + n) = 9n + 2logn >> 9n
占主导地位,所以时间复杂度是 O(n)
。如果我错了请纠正我!
class Solution(object):
def binarySearch(self,arr, l, r, x):
if r >= l:
mid = l + (r - l)/2
if arr[mid] == x: return mid
elif arr[mid] > x:return self.binarySearch(arr, l, mid-1, x)
else: return self.binarySearch(arr, mid + 1, r, x)
else:return -1
def twoSum(self, nums, target):
temp = [i for i in nums] # o(n) 1 time
nums = list(set(nums)) # o(n) 1 time
nums.sort() # o(nlogn) 1 time
for i in range(len(nums)): # o(n) 1 time
s = target - nums[i] # o(1) n times
idx_binary = self.binarySearch(nums, 0, len(nums)-1, s) # o(logn) > n times
if idx_binary > -1: # o(1) n times
idx = temp.index(s, temp.index(nums[i])+1) # o(n) > 1 time
return [temp.index(nums[i]), idx] # o(n) > 1 time
else:
return [temp.index(nums[i]), temp.index(s)] # o(n) > 1 time
你的化简有误,应该是9n + 2nlogn,其中nlogn占优,所以答案是O(nlogn)
我试着计算函数的时间复杂度"twoSum"并写下每行的成本和它执行的次数。最后,我制作了两个向量:一个用于成本,另一个用于频率,然后计算它们的内积。我得到以下信息:(n + n + nlogn + n + n + nlogn + n + n + n + n) = 9n + 2logn >> 9n
占主导地位,所以时间复杂度是 O(n)
。如果我错了请纠正我!
class Solution(object):
def binarySearch(self,arr, l, r, x):
if r >= l:
mid = l + (r - l)/2
if arr[mid] == x: return mid
elif arr[mid] > x:return self.binarySearch(arr, l, mid-1, x)
else: return self.binarySearch(arr, mid + 1, r, x)
else:return -1
def twoSum(self, nums, target):
temp = [i for i in nums] # o(n) 1 time
nums = list(set(nums)) # o(n) 1 time
nums.sort() # o(nlogn) 1 time
for i in range(len(nums)): # o(n) 1 time
s = target - nums[i] # o(1) n times
idx_binary = self.binarySearch(nums, 0, len(nums)-1, s) # o(logn) > n times
if idx_binary > -1: # o(1) n times
idx = temp.index(s, temp.index(nums[i])+1) # o(n) > 1 time
return [temp.index(nums[i]), idx] # o(n) > 1 time
else:
return [temp.index(nums[i]), temp.index(s)] # o(n) > 1 time
你的化简有误,应该是9n + 2nlogn,其中nlogn占优,所以答案是O(nlogn)