10000个数求第2大的数,不许用排序算法.

    int i = 0, j = 1;
    int MAX, max;
    while (j<sizeof(a)/sizeof(a[0]))
    {
        if (a[0] != a[j++])
        {
            j --;
            break;
        }
    }
    if (j == sizeof(a)/sizeof(a[0]))return -1;
    MAX=a[0]>a[j]?a[0]:a[j], max=a[0]<a[j]?a[0]:a[j];
    for (i=2; i < sizeof(a)/sizeof(a[0]; i ++)
    {
        if (max < a[i])
        {
            if (MAX < a[i])
            {
                max = MAX;
                MAX = a[i];
            }else
            {
                max = a[i];
            }
        }
    }
数组a为自己定义，max为所求的第二大值

确实没有想到