LeetCode(169. 求众数)

问题描述

给定一个大小为 n 的数组，找到其中的众数。众数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众数。

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

解决方案

class Solution:
    # two pass + dictionary
    def majorityElement1(self, nums):
        dic = {}
        for num in nums:
            dic[num] = dic.get(num, 0) + 1
        for num in nums:
            if dic[num] > len(nums)//2:
                return num

    # one pass + dictionary
    def majorityElement2(self, nums):
        dic = {}
        for num in nums:
            if num not in dic:
                dic[num] = 1
            if dic[num] > len(nums)//2:
                return num
            else:
                dic[num] += 1

    # TLE
    def majorityElement3(self, nums):
        for i in range(len(nums)):
            count = 0
            for j in range(len(nums)):
                if nums[j] == nums[i]:
                    count += 1
            if count > len(nums)//2:
                return nums[i]

    # Sotring
    def majorityElement4(self, nums):

        return sorted(nums)[len(nums)//2]

    # Bit manipulation
    def majorityElement5(self, nums):
        bit = [0]*32
        for num in nums:
            for j in range(32):
                bit[j] += num >> j & 1
        res = 0
        for i, val in enumerate(bit):
            if val > len(nums)//2:
                # if the 31th bit if 1,
                # it means it's a negative number
                if i == 31:
                    res = -((1 << 31)-res)
                else:
                    res |= 1 << i
        return res

    # Divide and Conquer
    def majorityElement6(self, nums):
        if not nums:
            return None
        if len(nums) == 1:
            return nums[0]
        a = self.majorityElement(nums[:len(nums)//2])
        b = self.majorityElement(nums[len(nums)//2:])
        if a == b:
            return a
        return [b, a][nums.count(a) > len(nums)//2]

    # the idea here is if a pair of elements from the
    # list is not the same, then delete both, the last
    # remaining element is the majority number
    def majorityElement(self, nums):
        count, cand = 0, 0
        for num in nums:
            if num == cand:
                count += 1
            elif count == 0:
                cand, count = num, 1
            else:
                count -= 1
        return cand