-
Notifications
You must be signed in to change notification settings - Fork 0
/
Copy path169-多数元素.py
58 lines (45 loc) · 1.29 KB
/
169-多数元素.py
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
'''
给定一个大小为 n 的数组,找到其中的多数元素。多数元素是指在数组中出现次数大于 ⌊ n/2 ⌋ 的元素。
你可以假设数组是非空的,并且给定的数组总是存在多数元素。
示例 1:
输入: [3,2,3]
输出: 3
示例 2:
输入: [2,2,1,1,1,2,2]
输出: 2
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/majority-element
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
'''
class Solution:
def majorityElement(self, nums) -> int:
count = len(nums)/2
nums_set = list(set(nums))
for i in nums_set:
sum = 0
for j in nums:
if j == i:
sum += 1
if sum >= count:
return i
else:
pass
def majorityElement_good(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
major = 0
count = 0
for n in nums:
if count == 0:
major = n
if n == major:
count = count + 1
else:
count = count - 1
return major
obj = Solution()
nums = [2, 2, 1, 1, 1, 2, 2]
print(obj.majorityElement_good(nums))