【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

 1 class Solution {

 2 public:

 3     int majorityElement(vector<int> &num) {

 4        

 5         int n=num.size();

 6         int result;          

 7         map<int,int> num2count;

 8  

 9         for(int i=0;i<n;i++)

10         {

11             if(num2count.find(num[i])!=num2count.end())

12             {

13                 num2count[num[i]]++;

14             }

15             else

16             {

17                 num2count[num[i]]=1;

18             }

19         }

20        

21         map<int,int>::iterator it=num2count.begin();

22         while(it!=num2count.end())

23         {

24             if(it->second>n/2)

25             {

26                 result=it->first;

27                 break;                 

28             }              

29             it++;

30         }

31         return result;

32     }

33 };

 

 

 

方法2，先排序，然后统计

 

 1 class Solution {

 2 public:

 3     int majorityElement(vector<int> &num) {

 4        

 5         sort(num.begin(),num.end());

 6         int result=num[0];

 7         int count=0;

 8         int max_count=1;

 9         int cur=num[0];

10        

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

12         {

13             if(num[i]==cur) 

14             {

15                 count++;

16             }

17             else

18             {

19  

20                 if(count>max_count)

21                 {

22                     result=cur;

23                     max_count=count;

24                 }

25                 cur=num[i];

26                 count=1;

27             }

28         }

29        

30         if(count>max_count)

31         {

32             result=cur;

33         }

34        

35         return result;

36  

37     }

38 };

 

 

方法3，采用随机选取元素的方法，进行统计：

 1 #define random(x) (rand()%x)

 2 class Solution {

 3 public:

 4     int majorityElement(vector<int> &num) {

 5        

 6         int result;

 7         while(1)

 8         {

 9             result=num[random(num.size())];

10             int count=0;

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

12             {

13                 if(num[i]==result)

14                 {

15                     count++;

16                 }

17             }

18             if(count>num.size()/2)

19             {

20                 break;

21             }

22         }

23         return result;

24     }

25 };

 

 

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 ).

 

 

 1 class Solution {

 2 public:

 3     int majorityElement(vector<int> &num) {      

 4        

 5         int count=0;

 6         int i=0;

 7         int result;

 8         while(i<num.size())

 9         {

10             if(count==0)

11             {

12                 result=num[i];

13                 count++;

14             }

15             else

16             {

17                 if(result==num[i])

18                 {

19                     count++;

20                 }

21                 else

22                 {

23                     count--;

24                 }

25             }

26             i++;

27         }       

28         return result;       

29     }

30 };