Majority Element II Solutions in C++

// Source : https://leetcode.com/problems/majority-element-ii/
// Author : Hao Chen
// Date   : 2015-07-03



class Solution {
public:

    //O(n) Space compexity
    vector<int> majorityElement01(vector<int>& nums) {
        vector<int> result;
        unordered_map<int, int> counts;
        int n = nums.size();
        for(auto item : nums){
            counts[item]++;
            if (counts[item] > n/3 ){
               result.push_back(item); 
               counts[item] = -n; // Tricky: make sure the item only can be put into result once.
            } 
        }
        return result;
    }
    //We know, there could be at most two numbers can be more than 1/3
    //so, same as Majority Element I problem, we can have two counters.
    vector<int> majorityElement02(vector<int>& nums) {
        if(nums.size()<=1) return nums;
        
        //the same algorithm as Majority Element I problem
        int majority1=0, majority2=0, cnt1=0, cnt2=0;
        for(auto item: nums) {
            if (cnt1 == 0 && majority2 != item ) {
                majority1 = item;
                cnt1 = 1;
            } else if (majority1 == item) {
                cnt1++;
            } else if (cnt2 == 0) {
                majority2 = item;
                cnt2 = 1;
            } else if (majority2 == item) {
                cnt2++;
            } else {
                cnt1--;
                cnt2--;
            }
        }
        //re-check it again, in case there has less than two numbers of majority
        cnt1 = cnt2 = 0;
        for (auto item : nums) {
            if (majority1 == item) cnt1++;
            else if (majority2 == item) cnt2++;
        }
        vector<int> result;
        if (cnt1 > nums.size()/3) result.push_back(majority1);
        if (cnt2 > nums.size()/3) result.push_back(majority2);
        return result;
        
    }
    
    vector<int> majorityElement(vector<int>& nums) {
        return majorityElement02(nums);
        return majorityElement01(nums);
    }
};