返回给定的二位整数数组中出现次数最多的元素；
 例如sz[2][3]={{3,1,3},{1,3,5}}
 打印3
求指教

#include<stdio.h>
#include<math.h>
main()
{
    int i,j,k,a,sz[2][3]={{3,1,3},{1,3,5}};
     int *q;
     int max=0;   
     q=sz;
         for(i=0;*(q+i)!='\0';i++)
         {
               for(k=0,j=i;*(q+j)!='\0';j++)
               {
                   if(*(q+j)==*(q+i))
                       k++;
               }
               if(k>max)
               { a=*(q+i);
                 max=k;
               }
     }
     printf("%d",a);
}

好像不对啊

怎么情况呢！！！

#include <stdio.h>
#define  N 2
#define  M 3
#define  K   N * M
#define  STRING  " element           times \n "
int main( )
{
    void arr_conv(int array[][M], int a2[K]);
    void elem_count(int array[K]);
    int  a1[N][M], a2[K];
    int  i, j;
    while (888)
    {
        printf("Please input a  two-dimensional array:");
        for (i = 0; i < N; i++)
            for (j = 0; j < M; j++)
                scanf("%d", &a1[i][j]);
            printf("\nNow original array a1[N][M] is :\n");
            for (i = 0; i < N; i++)
            {
                for (j = 0; j < M; j++)
                    printf("%-6d ", a1[i][j]);
                 printf("\n");
            }
            arr_conv(a1,a2);
            elem_count(a2);
    }
    return 0;
}

void  arr_conv(int array[][M], int a2[K])
{
    int i , j ;
    for (i = 0; i < N; i++)
    {
        for (j = 0; j < M; j++)
            a2[i * M + j] = array[i][j];
    }
    printf("a2[K] = {");
    for(i =  0; i < K - 1; i++)
    printf("%d,", a2[i]);
    printf("%d}.\n", a2[K - 1]);

}


void elem_count( int array[K] )
{   
    int i , j, n , m = 0, len = K, s ,a3[K];
    for (i = 0; i < len; i++)
    {
        for ( j = i + 1 , n = 0 ; j < len; j++)
          if (array[i] == array[j])
          {   
             for (s = j; s < len ; s++)   
              array[s] = array[s + 1];             // Delete the same number
              len--;
              j--;
              n++;
              a3[i] = n;
          }
          if (m < n)
              m = n;
          printf(STRING);
          printf("%6d  ----->      %d\n", array[i], n + 1 );
    }
    for (i = 0; i < len; i++)
    {
        if (a3[i] == m)
        printf("Element %d appeared mostly, it appeared %d times.\n", array[i], m + 1);
    }
}

[ 本帖最后由 有容就大 于 2011-12-18 01:28 编辑 ]

两次循环  用数组统计每个数出现的次数  ok