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majority-element-ii.py 1.43 KB
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Allen Liu 提交于 7年前 . add complexity
# Time: O(n)
# Space: O(1)
import collections
class Solution(object):
def majorityElement(self, nums):
"""
:type nums: List[int]
:rtype: List[int]
"""
k, n, cnts = 3, len(nums), collections.defaultdict(int)
for i in nums:
cnts[i] += 1
# Detecting k items in cnts, at least one of them must have exactly
# one in it. We will discard those k items by one for each.
# This action keeps the same mojority numbers in the remaining numbers.
# Because if x / n > 1 / k is true, then (x - 1) / (n - k) > 1 / k is also true.
if len(cnts) == k:
for j in cnts.keys():
cnts[j] -= 1
if cnts[j] == 0:
del cnts[j]
# Resets cnts for the following counting.
for i in cnts.keys():
cnts[i] = 0
# Counts the occurrence of each candidate integer.
for i in nums:
if i in cnts:
cnts[i] += 1
# Selects the integer which occurs > [n / k] times.
result = []
for i in cnts.keys():
if cnts[i] > n / k:
result.append(i)
return result
def majorityElement2(self, nums):
"""
:type nums: List[int]
:rtype: List[int]
"""
return [i[0] for i in collections.Counter(nums).items() if i[1] > len(nums) / 3]
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