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- nums = [1, 1, 2, 1, 3, 2, 3, 2]
-
- major1 = major2 = nums[0]
- count1 = count2 = 0
-
- # 对抗阶段
- for each in nums:
- if major1 == each:
- count1 += 1
- continue
-
- if major2 == each:
- count2 += 1
- continue
-
- if count1 == 0:
- major1 = each
- count1 = 1
- continue
-
- if count2 == 0:
- major2 = each
- count2 = 1
- continue
-
- count1 -= 1
- count2 -= 1
-
- # 统计阶段
- if nums.count(major1) > len(nums) / 3:
- print(major1)
- if nums.count(major2) > len(nums) / 3:
- print(major2)
复制代码
这里面,如何避免major1和major2相同的情况呢???
本帖最后由 傻眼貓咪 于 2022-1-22 09:23 编辑
试试这个: - nums = [1, 1, 2, 1, 3, 2, 3, 2]
- major1 = major2 = nums[0]
- count1 = count2 = 0
- for each in nums:
- if count1 == 0:
- major1 = each
- elif each == major1:
- count1 += 1
- else:
- count1 -= 1
-
- if count2 == 0 and each != major1:
- major2 = each
- elif each == major2:
- count2 += 1
- else:
- count2 -= 1
- if nums.count(major1) > len(nums)/3:
- print(major1)
- if nums.count(major2) > len(nums)/3:
- print(major2)
复制代码
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