Leetcode_Majority Element

Tag
Array/Divide and Conquer/Bit Manipulation
Difficulty
Easy
Description
Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times.
You may assume that the array is non-empty and the majority element always exist in the array.
Code

• Array：

Moore Voting Algorithm
每找出两个不同的element，即成对删除count–，否则count++
时间复杂度O(n)

class Solution:        # @param {integer[]} nums        # @return {integer}        def majorityElement(self, nums):            count=0            major_Element=0            for num in nums:                if count==0:                    major_Element=num                    count=count+1                else:                    if num==major_Element:                        count =count+1                    else:                        count = count-1            return major_Element

• Divide and Conquer
  将数组划分成两半，分别递归查找这部分的多数元素，如果相同，则为多数元素，否则为其中之一，判断其一即可
  时间复杂度O(nlogn)

class Solution(object):    def majority(self,nums,left,right):        if left == right:            return nums[left]        mid = (left+right)/2        lm = self.majority(nums,left,mid)        rm = self.majority(nums,mid+1,right)        if lm == rm:            return lm        if nums[left:right+1].count(lm) > nums[left:right+1].count(rm):            return lm        else:            return rm    def majorityElement(self, nums):        """        :type nums: List[int]        :rtype: int        """        return self.majority(nums,0,len(nums)-1)

• Bit Manipulation
  统计int的32位上每一位的众数，每一位的众数即组成了Majority Element的每一位。

class Solution {public:    int majorityElement(vector<int>& nums) {        int major = 0, n = nums.size();        for (int i = 0, mask = 1; i < 32; i++, mask <<= 1) {            int bitCounts = 0;            for (int j = 0; j < n; j++) {                if (nums[j] & mask) bitCounts++;                if (bitCounts > n / 2) {                    major |= mask;                    break;                }            }        }         return major;    } };