【leetcode】Majority Element

Majority Element

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.

方法1，采用一个map，存储每一个变量的数量，最后得出个数大于n/2的元素

class Solution {
public:
    int majorityElement(vector<int> &num) {
       
        int n=num.size();
        int result;          
        map<int,int> num2count;
 
        for(int i=0;i<n;i++)
        {
            if(num2count.find(num[i])!=num2count.end())
            {
                num2count[num[i]]++;
            }
            else
            {
                num2count[num[i]]=1;
            }
        }
       
        map<int,int>::iterator it=num2count.begin();
        while(it!=num2count.end())
        {
            if(it->second>n/2)
            {
                result=it->first;
                break;                 
            }              
            it++;
        }
        return result;
    }
};

 

 

方法2，先排序，然后统计

 

class Solution {
public:
    int majorityElement(vector<int> &num) {
       
        sort(num.begin(),num.end());
        int result=num[0];
        int count=0;
        int max_count=1;
        int cur=num[0];
       
        for(int i=0;i<num.size();i++)
        {
            if(num[i]==cur) 
            {
                count++;
            }
            else
            {
 
                if(count>max_count)
                {
                    result=cur;
                    max_count=count;
                }
                cur=num[i];
                count=1;
            }
        }
       
        if(count>max_count)
        {
            result=cur;
        }
       
        return result;
 
    }
};

 

方法3，采用随机选取元素的方法，进行统计：

#define random(x) (rand()%x)
class Solution {
public:
    int majorityElement(vector<int> &num) {
       
        int result;
        while(1)
        {
            result=num[random(num.size())];
            int count=0;
            for(int i=0;i<num.size();i++)
            {
                if(num[i]==result)
                {
                    count++;
                }
            }
            if(count>num.size()/2)
            {
                break;
            }
        }
        return result;
    }
};

 

多数投票算法：

过程：

Runtime: O( n ) — Moore voting algorithm: We maintain a current candidate and a counter initialized to 0. As we iterate the array, we look at the current element x: 

1. If the counter is 0, we set the current candidate to x and the counter to 1.
2. If the counter is not 0, we increment or decrement the counter based on whether x is the current candidate.

After one pass, the current candidate is the majority element. Runtime complexity = O( n ).

 

 

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