连接两个数组并将新元素插入数组

Question

所以我的目标是将两个数组连接在一起，并在数组的开头插入新数组的新大小。例如如果

Arr1 = {5, 1,2,3,4}    
and Arr2 = {5, 10, 20, 30, 40 };

NewArr = { 11, 5, 1, 2, 3, 4, 5, 10, 20, 30, 40 }

但是，当我运行我的代码时，我得到了

NewArr = {11 1 2 3 4 10 20 30 40 -12851 -12851}

看起来我丢失了两个数组中的第一个元素。这是我的代码。

short int* ConcatArray(short int* a1, short int* a2) {
    //size of array 1
    int n = *(a1 + 0);
    //size of array 2
    int m = *(a2 + 0);
    int i;
    int size = n + m;

    short int* newArr = new short int[n + m];

    // increase the size by 1
    size++;

    // shift elements forward 
    for (i = size; i >= 1; i--)
        *(newArr + i) = *(newArr + i - 1);

    // insert x at pos 
    *(newArr + 1 - 1) = size;

    // Create new Array 
    for (i = 1; i < n; i++) {
        *(newArr + i) = *(a1 + i);
    }
    
    for (int j = 1; j < m; j++, i++) {
        *(newArr + i) = *(a2 + j);
    }
    

    return newArr;
}

Answer 1

假设您希望避免手动编写代码以“1×1”复制数组元素，语言和标准库中有一些工具可以为您做到这一点。

这是现代 C++ 的实现方式：

short* ConcatArray(short* a1, short* a2)
{
    size_t len1 = a1[0];
    size_t len2 = a2[0];

    short* a3 = new short[len1 + len2 + 1]; // +1 for new length

    *a3 = (short)(len1 + len2 + 1);
    std::copy_n(a1, len1, a3+1);
    std::copy_n(a2, len2, a3 + 1 + len1);

    return a3;
}

旧的“C”方式（使用新的分配器）：

short* ConcatArray(short* a1, short* a2)
{
    size_t len1 = a1[0];
    size_t len2 = a2[0];

    short* a3 = new short[len1 + len2 + 1]; // +1 for new length

    *a3 = (short)(len1 + len2 + 1);
    memcpy(a3+1, a1, sizeof(*a1)*len1);
    memcpy(a3+1+len1, a2, sizeof(*a2)*len2);


    return a3;
}

Answer 2

如果您首先将值复制到正确的位置，则无需移动它们。这应该可以满足您的需求。

short int *ConcatArray(short int *a1, short int *a2)
{
    short int n = a1[0]; // size of array 1
    short int m = a2[0]; // size of array 2
    short int newSize = n + m + 1;

    short int *newArr = new short int[newSize];

    int i = 0; // current output position in newArr
    newArr[i++] = newSize;

    for (int j = 0; j < n; ++j, ++i) { // copy a1
        newArr[i] = a1[j];
    }

    for (int j = 0; j < m; ++j, ++i) { // copy a2
        newArr[i] = a2[j];
    }

    return newArr;
}

这是Compiler Explorer 上的代码。