Question

我一直在努力进行合并排序，到目前为止，我认为一切正常，除了我必须比较左右数组的部分。救命！（这是用于家庭作业，所以它必须与这个伪代码几乎相同：

/*MERGESORT(A, p, r)]
    if p < r
        q = (p+r)/2
        MERGESORT(A,p,q)
        MERGESORT(A,q + 1, r)
        MERGE(A,p,q,r)

MERGE(A,p,q,r)
    n1 = q - p + 1
    n2 = r - q
    let L[1...n1 + 1] and R[1...n2 + 1] be new arrays
    for i = 1 to n1
        L[i] = A[p + i - 1]
    for j = 1 to n2
        R[j] = A[q + j]
    L[n1 + 1] = INFINITY
    R[n2 + 1] = INFINITY
    i = 1
    j = 1
    for k = p to r
        if L[i] <= R[j]
            A[k] = L[i]
            i = i + 1
        else
            A[k] = R[j]
            j = j + 1*/

这些是合并排序函数注意：忽略整数类型，它什么也没做。

void CensusData::mergeSort(int type) {
    if(type == 0) //STOPPED FOR DEBUGGING
        MERGE_SORT(type, 0, data.size() - 1);
}

void CensusData::MERGE_SORT(int type, int p, int r){
    //int q;
    //cout << "data size " << data.size() << endl;
    std::cout << "MERGE_SORT START ///("<< p << ", " << r << ")" <<std::endl;
    if(p < r)
    {
        int q = (p + r)/2;
        MERGE_SORT(type, p, q);
        MERGE_SORT(type, q + 1, r);
        MERGE(type, p, q ,r);
    }
}

void CensusData::MERGE(int type, int p, int q, int r){
    if(type == 0)
    {
        std::cout << "MERGING WITH: (" << p << ", "<< q <<", " << r<< ")"<< std::endl;
        //int n1;
        //int n2;
        int n1 = q - p + 1;
        int n2 = r - q;
        cout << "N1: " << n1 <<" N2:" << n2 << endl;
        Record* L[n1 + 1];
        Record* R[n2 + 1];
        L[n1 + 1] = NULL;
        R[n2 + 1] = NULL;
        for(int i = 0; i < n1; i++)
        {
            //if (L[i] == NULL)
                //continue;
            cout << "P, I: " << p <<", "<< i<< endl;
            cout << "filling array L: " << data[p + i]->population << endl;
            L[i] = data[p + i];
            cout<< L[i]->population << endl;
        }
        //cout << "J: " << j << endl;
        for(int j = 0; j < n2; j++)
        {
            //if(R[j] == NULL)
                //continue;
            cout << "filling array R: " << data[q + j + 1]->population<<endl;
            R[j] = data[q + j + 1];
            cout << R[j]->population << endl;
        }
        //THIS IS WHERE I THINK THE PROBLEMS ARE OCCURING FROM VVVVVVVVVVV
        int i = 0;
        int j = 0;
        for(int k = p; k <= r; k++)
        {
            if(L[i]->population < R[j]->population)
            {
                cout << "TRUE" << endl;
                data[k] = L[i];
                i = i + 1;
            }
            else if (L[i]->population > R[j]->population)
            {
                cout << "FALSE" << endl;
                data[k] = R[j];
                j = j + 1;
            }
        }
           /*std::vector<Record*>::iterator it = data.begin();
   while (it != data.end()) {
      std::cout << *(*it)->city << ", "
                << *(*it)->state << ", "
                << (*it)->population << std::endl;
      it++;}*/

    }
}

输入:::

Vina, California, 237
San Francisco, California, 812826
Santa Fe, New Mexico, 68642

输出:::

Vina, California, 237
Santa Fe, New Mexico, 68642

Program received signal SIGSEGV, Segmentation fault.
0x00445cf7 in std::basic_ostream<char, std::char_traits<char> >& std::operator<< <char, std::char_traits<char>, std::allocator<char> >(std::basic_ostream<char, std::char_traits<char> >&, std::basic_string<char, std::char_traits<char>, std::allocator<char> > const&) ()
    at /usr/lib/gcc/i686-pc-cygwin/3.4.4/include/c++/iostream:77
77        static ios_base::Init __ioinit;

输入示例2 :::

Vina, California, 237
San Francisco, California, 812826
Santa Fe, New Mexico, 68642
Roseville, California, 1293
New York, New York, 283822
pieland, Conneticut, 283822

输出::::

Program received signal SIGSEGV, Segmentation fault.
0x004031ae in CensusData::MERGE (this=0x28aa40, type=0, p=0, q=2, r=5)
    at CensusDataSorts.cpp:105
105                 if(L[i]->population < R[j]->population)

Answer 1

这些语句错误（超出数组访问范围）

   L[n1 + 1] = NULL;
   R[n2 + 1] = NULL;

我认为你可能意味着

   L[n1] = NULL;
   R[n2] = NULL;

因为在伪代码中数组从1开始，但在代码数组中从零开始。我希望这种差异可能是你大多数问题的原因。

另一个探测器是你使用NULL来表示伪代码调用INFINITY的内容，但是当你进行比较时，你不会检查NULL。

所以

        if(L[i]->population < R[j]->population)

变为

        if ((L[i] != NULL && R[j] != NULL && L[i]->population < R[j]->population) ||
            (L[i] == NULL && R[j] != NULL))

下一个if语句的类似更改

        else if ((L[i] != NULL && R[j] != NULL && L[i]->population > R[j]->population) ||
            (L[i] != NULL && R[j] == NULL))

合并排序几乎与伪作品相同

1 个答案: