LeetCode229 多数元素 II

LeetCode229 多数元素 II

通过消除元素的方法确定候选, 即候选元素可以被消除\(\frac{n}{3}\)次.

class Solution:
    def majorityElement(self, nums: List[int]) -> List[int]:

        proposal_1, proposal_2, cnt_1, cnt_2, n = 0, 0, 0, 0, len(nums)

        for i in range(n):

            if cnt_1 > 0 and proposal_1 == nums[i]: cnt_1 += 1
            elif cnt_2 > 0 and proposal_2 == nums[i]: cnt_2 += 1
            elif cnt_1 == 0: proposal_1, cnt_1 = nums[i], 1
            elif cnt_2 == 0: proposal_2, cnt_2 = nums[i], 1
            else: cnt_1, cnt_2 = cnt_1 - 1, cnt_2 - 1
        
        sum_1, sum_2 = 0, 0
        for i in range(n):

            if cnt_1 > 0 and nums[i] == proposal_1: sum_1 += 1
            if cnt_2 > 0 and nums[i] == proposal_2: sum_2 += 1

        ans = []
        if sum_1 > n // 3: ans.append(proposal_1)
        if sum_2 > n // 3: ans.append(proposal_2)

        return ans

标签：cnt,nums,sum,元素,LeetCode229,II,elif,ans,proposal
来源： https://www.cnblogs.com/solvit/p/16505149.html