【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 };

 

posted @ 2014-12-24 18:44  H5开发技术  阅读(231)  评论(0)  编辑  收藏  举报