---
title: leetcode-169
date: 2020-03-23 23:12:57
tags:
categories: leetcode
---


### 169. 多数元素

[题目](https://leetcode-cn.com/problems/majority-element/)



<!--more-->



一开始的思路是遍历一遍整个列表，用一个字典去记录每个元素出现的次数，当次数大于 $\cfrac{n}{2}$ 时就可以得出结果。

```python
class Solution:
  def majorityElement(self, nums) -> int:
    d = {}
    l = len(nums)
    for n in nums:
      if not n in d:
        d[n] = 0
      d[n] = d[n] + 1
      if d[n] > l/2:
        return n
# 64 ms	15.1 MB
```

Python 也有专门计数的库，写起来更简单一点：

```python
class Solution:
  def majorityElement(self, nums):
    counts = collections.Counter(nums)
      return max(counts.keys(), key=counts.get)
# 44 ms	15.1 MB
```

由于要找的数出现次数大于 $\cfrac{n}{2}$，脑子里掠过一下[蒙特卡罗算法](https://zh.wikipedia.org/wiki/%E8%92%99%E5%9C%B0%E5%8D%A1%E7%BE%85%E6%96%B9%E6%B3%95)，后来在官方解答中也看到类似的思路了：

```python
class Solution:
    def majorityElement(self, nums):
        majority_count = len(nums)//2
        while True:
            candidate = random.choice(nums)
            if sum(1 for elem in nums if elem == candidate) > majority_count:
                return candidate

#作者：LeetCode-Solution
#链接：https://leetcode-cn.com/problems/majority-element/solution/duo-shu-yuan-su-by-leetcode-solution/
#来源：力扣（LeetCode）
#著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
```