64 lines
1.6 KiB
Markdown
64 lines
1.6 KiB
Markdown
---
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title: leetcode-169
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date: 2020-03-23 23:12:57
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tagstags:
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- leetcode:
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categories: leetcode
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---
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### 169. 多数元素
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[题目](https://leetcode-cn.com/problems/majority-element/)
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<!--more-->
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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。
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```python
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class Solution:
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def majorityElement(self, nums) -> int:
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d = {}
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l = len(nums)
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for n in nums:
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if not n in d:
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d[n] = 0
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d[n] = d[n] + 1
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if d[n] > l/2:
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return n
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# 64 ms 15.1 MB
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```
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Python 也有专门计数的库，写起来更简单一点：
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```python
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class Solution:
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def majorityElement(self, nums):
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counts = collections.Counter(nums)
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return max(counts.keys(), key=counts.get)
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# 44 ms 15.1 MB
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```
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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：
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```python
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class Solution:
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def majorityElement(self, nums):
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majority_count = len(nums)//2
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while True:
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candidate = random.choice(nums)
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if sum(1 for elem in nums if elem == candidate) > majority_count:
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return candidate
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#作者：LeetCode-Solution
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#链接：https://leetcode-cn.com/problems/majority-element/solution/duo-shu-yuan-su-by-leetcode-solution/
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#来源：力扣（LeetCode）
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#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。
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```
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