Problem statement:

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.

Solution one: sorting(AC)

The number of majority elements is greater than n/2. It must take n/2 if we sort the array no matter by ascending or descending order.

Time complexity is O(nlgn). 

class Solution {public:    int majorityElement(vector<int>& nums) {        sort(nums.begin(), nums.end());        return nums[nums.size() /2];    }};

Solution two: hash table(AC)

We can employ a hash table indexed by the element, corresponding to the number of the element.

loop from the beginning to end, Once the number of a element is greater than n / 2, return the element.

The worst case of time complexity is O(n), generally it is less than O(n). 

The worst case of space complexity is O(n), generally is less than O(n).

class Solution {public:    int majorityElement(vector<int>& nums) {        int size = nums.size();        unordered_map<int, int> table;        for(auto num : nums){            if((++table[num]) > size / 2){                return num;                }        }        return -1;    }};

Solution three: one-time loop(AC)

We use the philosophy of counteracting. 

One variable, candidate, represents the candidate, one, cnt,  represents the number of the candidate.

• If count == 0; candidate = num;
• else if candidate = num ---> count++;
• else if candidate != num ---> count--;

Time complexity is O(n). Space complexity is O(1).

class Solution {public:    int majorityElement(vector<int>& nums) {        int candidate = 0;        int cnt = 0;        for(auto num : nums){            if(cnt == 0){                candidate = num;                cnt++;            } else if (candidate == num) {                cnt++;            } else {                cnt--;            }        }        return candidate;    }};

169. Majority Element