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问对链表数组进行合并排序
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Stack Overflow用户

提问于 2019-03-01 00:30:24

回答 1查看 187关注 0票数 0

我正在尝试合并、排序一个数组的每个元素，它是一个链表的头。

//for each element in array
for(int i=0; i<MAX_SIZE; i++){

    //NOTE: this current element (i) holds a linked list...sort it

    //merge function should accept head of linked list which is this element
    mergeSort(&inventory[i]);

}

下面是我的合并排序函数，它递归地调用自己...

void mergeSort(data** A){

data* head = *A;  //head is begining of A
data* r;
data* l;


//base case ... if linked list length is 0 or 1
if(head == nullptr || head->next == nullptr){
    return;
}

//divide linked list into halves
split(head, &r, &l);

//r holds first half of linked list
//l holds left second half of linked list

mergeSort(&r);
mergeSort(&l);
*A = merge(r,l);  //merge into linked list A
}

下面是split算法。

void split(data* source, data** R, data** L){ //source is head of this linked list

data* slow = source;
data* fast = source->next;

while(fast->next->next){ //while fast has two nodes after it  
    fast= fast->next;
    slow = slow->next;
}


*L = source;
*R = slow->next;
slow->next = nullptr;
}

最后我将这些段合并到链表A中，下面的函数返回链表的头部并设置为A…

data* merge(data* R, data* L){

data* result = nullptr;


//base cases
if(R == nullptr){
    return R;
}
else if(L == nullptr){
    return L;
}

if(R->pnum <= L->pnum){
    result = R;
    result->next = merge(R->next, L);
}
else{
    result = L;
    result->next = merge(L, R->next);
}

return result;

}

当我尝试用一个简单的打印函数打印整个数组时，代码成功运行，但没有打印出来，我认为合并代码永远不会停止运行。终端中没有显示错误(使用xcode的即时消息)。

对如何解决这个问题有什么建议吗？谢谢!

linked-list

mergesort

c++

recursion

回答 1

Stack Overflow用户

发布于 2019-03-01 05:04:25

请尝试使用此合并排序代码。

/* C program for Merge Sort */
#include<stdlib.h> 
#include<stdio.h> 

// Merges two subarrays of arr[]. 
// First subarray is arr[l..m] 
// Second subarray is arr[m+1..r] 
void merge(int arr[], int l, int m, int r) 
{ 
    int i, j, k; 
    int n1 = m - l + 1; 
    int n2 = r - m; 

    /* create temp arrays */
    int L[n1], R[n2]; 

    /* Copy data to temp arrays L[] and R[] */
    for (i = 0; i < n1; i++) 
        L[i] = arr[l + i]; 
    for (j = 0; j < n2; j++) 
        R[j] = arr[m + 1+ j]; 

    /* Merge the temp arrays back into arr[l..r]*/
    i = 0; // Initial index of first subarray 
    j = 0; // Initial index of second subarray 
    k = l; // Initial index of merged subarray 
    while (i < n1 && j < n2) 
    { 
        if (L[i] <= R[j]) 
        { 
            arr[k] = L[i]; 
            i++; 
        } 
        else
        { 
            arr[k] = R[j]; 
            j++; 
        } 
        k++; 
    } 

    /* Copy the remaining elements of L[], if there 
    are any */
    while (i < n1) 
    { 
        arr[k] = L[i]; 
        i++; 
        k++; 
    } 

    /* Copy the remaining elements of R[], if there 
    are any */
    while (j < n2) 
    { 
        arr[k] = R[j]; 
        j++; 
        k++; 
    } 
} 

/* l is for left index and r is right index of the 
sub-array of arr to be sorted */
void mergeSort(int arr[], int l, int r) 
{ 
    if (l < r) 
    { 
        // Same as (l+r)/2, but avoids overflow for 
        // large l and h 
        int m = l+(r-l)/2; 

        // Sort first and second halves 
        mergeSort(arr, l, m); 
        mergeSort(arr, m+1, r); 

        merge(arr, l, m, r); 
    } 
} 

/* UTILITY FUNCTIONS */
/* Function to print an array */
void printArray(int A[], int size) 
{ 
    int i; 
    for (i=0; i < size; i++) 
        printf("%d ", A[i]); 
    printf("\n"); 
} 

/* Driver program to test above functions */
int main() 
{ 
    int arr[] = {12, 11, 13, 5, 6, 7}; 
    int arr_size = sizeof(arr)/sizeof(arr[0]); 

    printf("Given array is \n"); 
    printArray(arr, arr_size); 

    mergeSort(arr, 0, arr_size - 1); 

    printf("\nSorted array is \n"); 
    printArray(arr, arr_size); 
    return 0; 
}