LeetCode----Majority Element

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.

分析：

簡單題。思路1：使用字典。思路2：使用Counter。思路3：使用Moore's voting algorithm。

解釋下思路3：

Runtime: O(n) — Moore voting algorithm: We maintain a current candidate and a counter initialized to 0. As we iterate the array, we look at the current element x:

1. If the counter is 0, we set the current candidate to x and the counter to 1.
2. If the counter is not 0, we increment or decrement the counter based on whether x is the current candidate.

代碼1：

class Solution(object):
    def majorityElement(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        nd = {}
        nums_len = len(nums)
        half_len = nums_len / 2
        if nums_len == 1:
            return nums[0]
        for i in nums:
            if i in nd:
                if nd[i] >= half_len:
                    return i
                nd[i] += 1
            else:
                nd[i] = 1
           

代碼2：

class Solution(object):
    def majorityElement(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        from collections import Counter
        return Counter(nums).most_common(1)[0][0]
           

代碼3：

class Solution(object):
    def majorityElement(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        n = 1
        cur = nums[0]
        for i in nums[1:]:
            if i == cur:
                n += 1
            else:
                if n == 0:
                    cur = i
                else:
                    n -= 1
        return cur