已排好序的整形數組的整合

例如已知數組a前半部分a[0,mid - 1],後半部分a[mid,num-1]，現前半部分和後半部分均已排好序。

要求：實現a數組的從小到大排序。空間複雜度爲O(1).(百度筆試)

#include <stdio.h>

void MergeIntData(int *a, const unsigned int num, const unsigned int mid)
{
    if(mid < 0 || num < 0 || NULL == a || a[mid - 1] <= a[ mid ])
        return;

    unsigned int i,temp, low = 0, high = 0;

    if(mid <= num / 2)
    {
        while(low < mid)
        {
            while(a[low] < a[mid + high])//不能加等號,否則如果有連續相等在一起的數據，會出現錯誤
            {
                ++ low;
            }

            while(a[mid + high + 1] < a[low])
            {
                ++ high;

                if(high  >= num - mid)
                {
                    high = num - mid - 1;
                    break;
                }
            }

            temp = a[low];
            for(i = low; i < mid + high; ++ i)
            {
                a[i] =  a[i + 1];
            }
            a[mid + high] = temp;
        }
    }
    else
    {
        while(high < num - mid)
        {
            while(a[low] < a[mid + high])
            {
                ++ low;
            }

            temp = a[mid + high];
            for(i = mid + high; i > low; -- i)
            {
                a[i] =  a[i - 1];
            }
            a[low] = temp;

            ++ high;
        }
    }
}


int main()
{

//  int a[] = {0,2,4,6,8,9,10,11,12,14,16,18,1,3,5,7};
//  int a[] = {0,2,4,6,8,8,9,9,1000,1,3,5,7,10,11,12,14,16,18,20,21,23,25,45,68};
//  int a[] = {100,1,3,5,7,10,11,12,14,16,18,20,21,23,25,45,68};
//  int a[] = {8,8,8,9,9,1,5,8,9,10,12,13,14};
//  int a[] = {1,2,3,4,5,6,7,8,9,2,4,5,7,8,9,10};
//  int a[] = {0,1,2,3,4,5,6,7,8,9,9,17,100,5,5,5,5,9,10,10000};

    int a[] = {1,2,3,4,5,6,7,100,1000,2,3,5,6,8,1001};
    int len = sizeof(a) / sizeof(int);
    int mid =  9;

    MergeIntData(a, len, mid);

    for(int i = 0; i < len; ++ i)
    {
        printf("%d  ", a[i]);
    }

    printf("\n");
    return 0;
}