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