数组中出现次数超过一半的数字

class Solution {
public:
    int MoreThanHalfNum_Solution(vector<int> numbers) {
    	if(numbers.size()==0)	return 0;
        sort(numbers.begin(),numbers.end());
    	int min=numbers[0];
        int max=numbers[numbers.size()-1];
        int k=max-min;
        int t;
        vector<int> book(k+1,0);
        
        for(int i=0;i<numbers.size();i++) {
        	book[numbers[i]]++;	
        }
        
        for(int j=0;j<k;j++) {
            if(book[j]>(numbers.size()/2)) {
                t=book[j];
                break;
            }
                
        }
        
      return t;
    }          
};

class Solution {
public:
    int MoreThanHalfNum_Solution(vector<int> numbers) {
    	int n=numbers.size();
        if(n==0)	return 0;
        int num=numbers[0],count=1;
        for(int i=1;i<n;i++) {
            if(numbers[i]==num) count++;
            else count--;
            if(count==0) {
                num=numbers[i];
                count=1;
            }
        }
        
        count=0;
        for(int i=0;i<n;i++) {if(numbers[i]== num) count++;}
        if(count * 2 >n) return num;
        return 0;
    }           
};

class Solution {
public:
    int MoreThanHalfNum_Solution(vector<int> numbers) 
    {
        // 因为用到了sort，时间复杂度O(NlogN)，并非最优
        if(numbers.empty()) return 0;
         
        sort(numbers.begin(),numbers.end()); // 排序，取数组中间那个数
        int middle = numbers[numbers.size()/2];
         
        int count=0; // 出现次数
        for(int i=0;i<numbers.size();++i)
        {
            if(numbers[i]==middle) ++count;
        }
         
        return (count>numbers.size()/2) ? middle :  0;
    }
};

class Solution {
public:
    int MoreThanHalfNum_Solution(vector<int> numbers) 
    {
        if(numbers.empty()) return 0;
         
        // 遍历每个元素，并记录次数；若与前一个元素相同，则次数加1，否则次数减1
        int result = numbers[0]; 
        int times = 1; // 次数
         
        for(int i=1;i<numbers.size();++i)
        {
            if(times == 0)
            {
                // 更新result的值为当前元素，并置次数为1
                result = numbers[i];
                times = 1;
            }
            else if(numbers[i] == result)
            {
                ++times; // 相同则加1
            }
            else
            {
                --times; // 不同则减1                
            }
        }
         
        // 判断result是否符合条件，即出现次数大于数组长度的一半
        times = 0;
        for(int i=0;i<numbers.size();++i)
        {
            if(numbers[i] == result) ++times;
        }
         
        return (times > numbers.size()/2) ? result : 0;
    }
};