[Leetcode]Majority Element II

Given an integer array of size n, find all elements that appear more than⌊ n/3 ⌋ times. The algorithm should run in linear time and in O(1) space.

class Solution {public:    /*alorithm: hash soluton      1)count elements occurence usin hash table      2)iterate hash table to check whether it > n/3 floor       time O(n) space O(n)    */    vector<int> majorityElement(vector<int>& nums) {                vector<int>ret;                unordered_map<int,int>hash;                int n = nums.size();                //count                for(int i = 0;i < n;i++){                    hash[nums[i]]++;                }                //check n/3 number                for(auto it = hash.begin();it != hash.end();it++){                    if(it->second > n/3)                        ret.push_back(it->first);                }                                return ret;    }};

class Solution {public:    /*algorithm:voting algorithm    time O(n) space O(1)    */    vector<int> majorityElement(vector<int>& nums) {             vector<int>ret;            int n = nums.size();            int n1,n2,c1,c2;            c1 = c2 = 0;            for(int i = 0;i < n;i++){                if(n1 == nums[i])c1++;                else if(n2 == nums[i])c2++;                else if(c1 == 0)n1 = nums[i],c1++;                else if(c2 == 0)n2 = nums[i],c2++;                else c1--,c2--;            }            c1 = c2 = 0;            for(int i = 0;i < n;i++){                if(n1 == nums[i])c1++;                if(n2 == nums[i])c2++;            }            if(c1 > n/3)ret.push_back(n1);            if(c2 > n/3)ret.push_back(n2);            return ret;    }};