AC python 1 line nlogn time

  • 12
    class Solution:
        # @param num, a list of integers
        # @return an integer
        def majorityElement(self, num):
            return sorted(num)[len(num)/2]

  • 0

    I like this. I completely didn't realize that the major element will be in the n/2's slot. Instead of getting the length of each repeated element, simply go for the n/2 spot.

  • 0

    Very nice code!

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