【每日一题Leetcode】-169. Majority Element

最新推荐文章于 2019-06-24 21:33:33 发布

最新推荐文章于 2019-06-24 21:33:33 发布 · 174 阅读

Leetcode刷刷刷专栏收录该内容

21 篇文章

订阅专栏

本文介绍了一种寻找数组中出现次数超过一半的元素（多数元素）的算法。通过遍历列表并使用字典记录每个元素的出现次数，我们可以找到多数元素。此外，还提供了一种更高效的解决方案，即在遍历过程中，一旦发现某个元素的计数超过数组长度的一半，立即返回该元素。

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.

Example 1:

Input: [3,2,3]
Output: 3

Example 2:

Input: [2,2,1,1,1,2,2]
Output: 2

这个题目比较简单，题目可以理解为求出现次数最多的元素。解题思路：遍历整个列表，保存各元素及计数到字典。为了少遍历一次字典，增加一个元组用来保存次数最多元素的值和次数。

class Solution(object):
    def majorityElement(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        
        counts = {}
        major = (0,0) #(number,count)
        for i in nums:
            if i in counts:
                counts[i] += 1
            else:
                counts[i] = 1
            
            if counts[i] > major[1]:
                major = (i,counts[i])
                
        return major[0]

44 / 44 test cases passed.	Status: Accepted
Runtime: 48 ms

在论坛看到一种时间复杂度低一些的思路，每遍历一个元素，更新计数后，将计数与字典长度的一半做比较。大于这个值直接返回该元素。我觉得这个方法更好

class Solution(object):
    def majorityElement(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        len2 = len(nums)//2
        counts = {}
        #major = (0,0) #(number,count)
        for i in nums:
            if i in counts:
                counts[i] += 1
            else:
                counts[i] = 1
            
            if counts[i] > len2:
                return i
                
        return

参考资料：https://leetcode.com/problems/majority-element/discuss/51712/Python-different-solutions-(dictionary-bit-manipulation-sorting-divide-and-conquer-brute-force-etc).