排序01

假设一整型数组存在若干正数和负数，现在通过某种算法使得该数组的所有负数在在正数的左边，且保证负数和正数间元素相对位置不变。时空复杂度要求分别为o(n), o(1).

//算法分开正负数
void insertSort(int*A ,int size)
{ 
    int minus=-1,plus=-1;
    int tmp=0;
    for(int i=0;i<size;i++)
    {
        if(minus==-1)// means start sorting or has done one sorting.
        {
            if(A[i]<0&& plus>=0)// means plus number has been found and finds another minus number.
            {
                minus=i;
            }
            
            if(A[i]>0&& plus<0)// means find first plus number.
            {
                plus=i;
            }
        }

        if(minus>=0&& plus>=0) // replace [plus, ..., minus] to [minus, plus, ...]
        {
            tmp=A[plus];
            A[plus++] = A[minus];
            for (int k = minus; k> plus; k--) 
            {
                 A[k] = A[k -1];
            }
            A[plus]=tmp;
            minus=-1;
        }
    }
}

Note:

1 plus is increased 1 each time a plus number is found.