[LeetCode] Majority Element

Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times.

You may assume that the array is non-empty and the majority element always exist in the array.

Credits:

Special thanks to @ts  for adding this problem and creating all test cases.

其实这道题求的是数组中出现频率超过 n / 2的元素，之前看过一种解题方法：用两个标记量，一个标记元素的值，一个标记出现次数，遍历数组，如果新元素与标记元素相等，则标记次数递增，否则标记次数减一，当次数为0时，将新元素赋值为标记元素，出现次数重置为1，最后得到的标记元素就是出现次数 > n / 2的元素。这种解法的缺点就是必须要求数组中有出现频率超过 n /2。

class Solution:
	# @param {integer[]} nums
	# @return {integer}
	def majorityElement(self, nums):
		if len(nums) < 1:
			return -1
			
		maj_val = nums[0]
		maj_cnt = 1

		for i in range(len(nums)):
			if nums[i] == maj_val:
				maj_cnt += 1
			else:
				maj_cnt -= 1
				if 0 == maj_cnt:
					maj_val = nums[i]
					maj_cnt = 1
		return maj_val

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