229. Majority Element II 【M】【52】

原创于 2016-05-15 15:28:01 发布 · 351 阅读

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本文介绍了一种线性时间和O(1)空间复杂度的算法来找出数组中出现次数超过n/3的元素。该算法使用抵消方法进行处理，并提供了Python实现。

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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还是用抵消的办法

开始的参数是0.5，通过了

后来改成1 ，也通过了……

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