[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;
}
};