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top-k-frequent-elements.py
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top-k-frequent-elements.py
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# Time: O(n) ~ O(n^2), O(n) on average.
# Space: O(n)
# Given a non-empty array of integers,
# return the k most frequent elements.
#
# For example,
# Given [1,1,1,2,2,3] and k = 2, return [1,2].
#
# Note:
# You may assume k is always valid,
# 1 <= k <= number of unique elements.
# Your algorithm's time complexity must be better
# than O(n log n), where n is the array's size.
from random import randint
class Solution(object):
def topKFrequent(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: List[int]
"""
counts = collections.defaultdict(int)
for i in nums:
counts[i] += 1
p = []
for key, val in counts.iteritems():
p.append((val, key))
self.kthElement(p, k);
result = []
for i in xrange(k):
result.append(p[i][1])
return result
def kthElement(self, nums, k):
def PartitionAroundPivot(left, right, pivot_idx, nums):
pivot_value = nums[pivot_idx][0]
new_pivot_idx = left
nums[pivot_idx], nums[right] = nums[right], nums[pivot_idx]
for i in xrange(left, right):
if nums[i][0] > pivot_value:
nums[i], nums[new_pivot_idx] = nums[new_pivot_idx], nums[i]
new_pivot_idx += 1
nums[right], nums[new_pivot_idx] = nums[new_pivot_idx], nums[right]
return new_pivot_idx
left, right = 0, len(nums) - 1
while left <= right:
pivot_idx = randint(left, right)
new_pivot_idx = PartitionAroundPivot(left, right, pivot_idx, nums)
if new_pivot_idx == k - 1:
return
elif new_pivot_idx > k - 1:
right = new_pivot_idx - 1
else: # new_pivot_idx < k - 1.
left = new_pivot_idx + 1
# Time: O(nlogk)
# Space: O(n)
class Solution2(object):
def topKFrequent(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: List[int]
"""
return [key for key, _ in collections.Counter(nums).most_common(k)]