leetcode - majority elements

leetcode - majority elements

给定size 为n的数组，查找出主元素，就是出现次数大于n/2次的元素。

1.

 int majorityElement(vector<int> &num) { 

std::map<int, int> im; 

for (int i = 0; i < num.size(); ++i){ 

map<int, int>::iterator it = im.find(num[i]);

 if (it == im.end()) 

{ im[num[i]] = 1; } 

else 

{ im[num[i]]++; } 

if (im[num[i]] > num.size()/2) 

{ return num[i]; } }

 return 0;}

2.

class Solution {public:    int majorityElement(vector<int>& nums) {        int ret = nums[0];        int count = 1;        for(int i = 1; i < nums.size(); i ++)        {            if(nums[i] == ret)                count ++;            else            {                count --;                if(count == 0)                {                    ret = nums[i];                    count = 1;                }            }        }        return ret;    }};

3. Moore's voting algorithm

int majorityElement(vector<int> &num) {    

int majorityIndex = 0;    

for (int count = 1, i = 1; i < num.size(); i++) 

{        

num[majorityIndex] == num[i] ? count++ : count--;        

if (count == 0) {           

 majorityIndex = i;            

count = 1;      

  }    

}    

return num[majorityIndex];

}

4.

class Solution { 2 public: 3     int majorityElement(vector<int> &num) { 4  5         int count = 0; 6          7         for(;;) { 8             if(num.size() == 1) 9                 return num[0];10             else    {11                 int i = rand() % (num.size() - 1);12                 for(int j = 0; j < num.size(); j++) {13                     if(num[j] == num[i])14                         count++;15                 }16                 if(count > (num.size() / 2))17                     return num[i];18                 else    {19                     count = 0;20                     continue;21                 }22             }23         }24     }
5.
 int my_cmp(int a,int b)  
{      
return a > b; 
 }    
int Solution::majorityElement(vector<int> &num) 
 {      
sort(num.begin(),num.end(),my_cmp);     
 return num[num.size()/2];
  }