using System;
using System.Collections.Generic;
using System.Text;
using System.Collections;

namespace 算法大挑战
{
    class Program
    {
        public ArrayList arr = new ArrayList();
        private int MajorityElement(int[] array,int n)
        {

            int count = 0;
            int num = 0;
            for (int i = 0; i < n; i++)
            {
                for (int j = i + 1; j < n; j++)
                {
                    if ((array[i]^array[j])==0)
                    {
                        count++;
                        if ((double)(count / 8) > 0.5)
                        {
                            num = array[i];
                        }
                    }
                }
            }
            return num;
        }

        static void Main(string[] args)
        {
            int[] array ={2,2,3,4,2,5,3,2,2,2};
            int n = array.Length;
            Program p = new Program();
            int num = p.MajorityElement(array,n);
            Console.WriteLine("{0}",num);
            Console.ReadKey();
        }
    }
}

------解决方案--------------------
如过数组是有序的，我的思路如下：
C# code

int MajorityElement(int array[],int n)
{
int NoMajorityElement=-1;
int len=array[].length;
int n=array[0];
int num=0;
for(int i=1;i<len;i++)
{
if(array[i]=n)
{
num++;
float fl=num/len;
if(fl>= o.51)
{
retrun n;
break;
}
}
else
{
n=array[i];
num=0;
}
if(i=m)
{
return NoMajorityElement;
}
}

------解决方案--------------------
让我走走诡异路线：

C# code


using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Text.RegularExpressions;

namespace ConsoleApplication2
{
    class Program
    {
        static int MajorityElement(int[] array, int n)
        {
            string s = "";
            Array.Sort(array);
            for (int i = 0; i < array.Length; i++)
                s += array[i];
            int iTemp = 0;
            int iMax = 0;
            MatchCollection mc = Regex.Matches(s, @"(\d)\1+");
            foreach (Match m in mc)
            {
                iTemp = m.Value.Length;
                if (iMax < iTemp)
                {
                    iMax = iTemp;
                    n = Convert.ToInt32(m.Value[0].ToString());
                }
            }
            if (iMax > array.Length / 2)
                return n;
            return -1;
        }

        static void Main(string[] args)
        {
            int[] array = { 2, 2, 3, 4, 2, 5, 3, 2, 2, 2 };
            int n = array.Length;
            int num = MajorityElement(array, n);
            Console.WriteLine("{0}", num);
            Console.ReadKey();
        }
    }
}

------解决方案--------------------
public find(int[] arr){   //只有当某一元素在数组中的绝对数超过50%，函数才能返回最多数。        int ele=0;
       int rud=0;
      
       foreach(int n in arr)
       {
           if(rud==0)
           {
               ele=n;
               rud=1;
           }
           else if(ele==n)
           {
               rud++;
           }
           else  
           {
               rud--;
           }
       }
      
       if(rud>0) return ele
       return 0;
}

------解决方案--------------------