在O（n）时间复杂度内找到出现超过一半的数

#include<iostream>
using namespace std;


bool solver(const int a[],const int n, int & num)
{
    if(NULL == a || 0>= n) return false; ////注意，是小写~

    int count = 0;
    int com = a[0];

    for(int i = 1;i<n;i++)
    {
     if(0 == count)
     {
         com = a[i];
         count++;
     }
     else{
         if(a[i] == com)
         count++;
         else
         count--;
       }
    }
    num = com;
    return true;
}

int main()
{
    int a[] = {1,2,3,3,3,3,3,3,4};
    int temp;
    solver(a,9,temp);
    cout<<temp;
}