一个从小到大排列的数组，这个数组中的一个平台就是连续的一段值相同的元素。例如：122333445中22，333等都是平台，333为最长平台。


我的程序 有错 没法运行

第15行有错，怎么错啦？

我这个思路可以吗？   还可以简化吗？


#include<stdio.h>
#define SIZE 1000
int maxLen(int , int);

main()
{
    int a[SIZE];
    int n, i;
    printf("Please input the number of intergers:");
    scanf("%d", &n);   //输入N个数
   
    for(i=0; i<=n-1; i++)  //依次读取N个数
        scanf("%d", a[i]);
   
    i=maxLen(a[], n);   //调用函数，求最长平台  怎么错的 （[Error]cpp:15: error: expected primary-expression before ']' token）
    printf("%d", i);
   
    return 0;
}                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                               


int maxLen(int a[], int n)
{
    int b[1000]={0};     //b[]数组做辅助
    int j, pass, hold;
   
    for(j=0; j<=n-1; j++){
   
        if(a[j]==a[j+1])     //若相同，b[这个数]加一
            b[a[j]]++;         
        
    }
   
    for(pass=0; pass<=n-1; pass++){    //泡沫排序，把最大的数放在b[0]
        for(j=0; j<=n-2; j++){
            if(b[j]<b[j-1]){
                hold=b[j];
                b[j]=b[j+1];
                b[j+1]=hold;
            }
        }
    }
   
    return b[0];
   
}

仍一枚硬币，正面和反面向上的概率都是0.5，但多次仍硬币，总会出现连续多次同一面的情况。我们把每次的实验结果记录下来，形成一个串（0，1分别代表正面、反面向上）。下面的程序统计出0或1最大连续出现的次数。试完善之。
    char* p = "100111000100101010001010111100000000100001111111000000";
    int len = strlen(p);
    int max_k = 1;
    int k = 1;
    for(int i=1; i<len; i++)
    {
        if(p[i]==p[i-1])
            k++;
        else
        {
            if(k>max_k) max_k = k；
            k = 1;
        }
    }
    printf("%d\n", max_k);

int maxLen(int[] , int);

在声明的时候就错了。。

声明改成  int maxLen(int[] , int);

但是15行怎么还是错的？

题目要求必须要调用函数

#include<stdio.h>
#define SIZE 1000
int maxLen(int[] , int);

main()
{
    int a[SIZE];
    int n, i;
    printf("Please input the number of intergers:");
    scanf("%d", &n);   //输入N个数
   
    for(i=0; i<=n-1; i++)  //依次读取N个数
        scanf("%d", a[i]);
   
    i=maxLen(a, n);   //调用函数，求最长平台  怎么错的 （[Error]cpp:15: error: expected primary-expression before ']' token）
    printf("%d", i);
   
    return 0;
}                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                               


int maxLen(int a[], int n)
{
    int b[1000]={0};     //b[]数组做辅助
    int j, pass, hold;
   
    for(j=0; j<=n-1; j++){
   
        if(a[j]==a[j+1])     //若相同，b[这个数]加一
            b[a[j]]++;         
        
    }
   
    for(pass=0; pass<=n-1; pass++){    //泡沫排序，把最大的数放在b[0]
        for(j=0; j<=n-2; j++){
            if(b[j]<b[j-1]){
                hold=b[j];
                b[j]=b[j+1];
                b[j+1]=hold;
            }
        }
    }
   
    return b[0];
   
}


这个可以开始运行了，没错了，但是运行一办还是不行