leetcode之Majority Element II
2015-10-18 16:34
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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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