leetcode之Majority Element II
来源:互联网 发布:林少华 知乎 编辑:程序博客网 时间:2024/05/21 17:39
majority element 2比1的不同在于1:不再是一定存在元素;2:由n/2变为n/3。1的可以排序直接查中间的元素来做,2就不行了。祭出来Counter就十分轻松啦。代码如下:
class Solution(object): def majorityElement(self, nums): """ :type nums: List[int] :rtype: List[int] """ from collections import Counter if nums == []: return [] a = Counter(nums) b = [] for i in a.keys(): if a[i] > len(nums) / 3: b.append(i) return b实际上可以把一开始的空集判断也是正确的。
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