Given an integer array of size?n, find all elements that appear more than?? n/3 ??times. The algorithm should run in linear time and in O(1) space.

Hint:

How many majority elements could it possibly have?

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還是用抵消的辦法
開(kāi)始的參數(shù)是0.5，通過(guò)了
后來(lái)改成1 ，也通過(guò)了……

class Solution(object):def majorityElement(self, nums):a = '' #nums[0]b = '' #nums[1]ta = 0tb = 0l = len(nums)if l == 2:return list(set(nums))times = l/3for i in nums[:]:if i == a:ta += 1elif i == b:tb += 1elif ta <= 0:a = ita = 1elif tb <= 0:b = itb = 1else:ta -= 1tb -= 1res = []if nums.count(a) > times:res += a,if a != b and nums.count(b) > times:res += b,return res

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