两个单链表之间相连的代码

可以按照以下步骤实现：

1. 找到单链表a中第i-1个元素，记为pre，以及第i+len-1个元素，记为next；
2. 将pre的next指针指向next；
3. 将单链表a的头节点插入到单链表b的第j个元素之前，具体实现可以参考单链表的插入操作。

具体代码如下：

// 找到单链表a中第i-1个元素和第i+len-1个元素
ListNode* pre = heada;
for (int k = 1; k < i; k++) {
    pre = pre->next;
}
ListNode* next = pre;
for (int k = ; k < len; k++) {
    next = next->next;
}

// 删除单链表a中第i个元素到第i+len-1个元素
pre->next = next;

// 将单链表a插入到单链表b的第j个元素之前
ListNode* p = headb;
for (int k = 1; k < j; k++) {
    p = p->next;
}
ListNode* temp = p->next;
p->next = heada;
while (heada->next != NULL) {
    heada = heada->next;
}
heada->next = temp;

需要注意的是，上述代码中的单链表节点类型为ListNode，需要根据实际情况进行修改。

相关问题

连接两个单链表的c语言代码

以下是连接两个单链表的C语言代码：

#include <stdio.h>
#include <stdlib.h>

struct Node {
    int data;
    struct Node* next;
};

void printList(struct Node* node) {
    while (node != NULL) {
        printf("%d ", node->data);
        node = node->next;
    }
}

void push(struct Node** head_ref, int new_data) {
    struct Node* new_node = (struct Node*)malloc(sizeof(struct Node));
    new_node->data = new_data;
    new_node->next = (*head_ref);
    (*head_ref) = new_node;
}

struct Node* mergeLists(struct Node* head1, struct Node* head2) {
    struct Node* mergedList = NULL;
    struct Node** lastPtrRef = &mergedList;

    while (1) {
        if (head1 == NULL) {
            *lastPtrRef = head2;
            break;
        }
        else if (head2 == NULL) {
            *lastPtrRef = head1;
            break;
        }

        if (head1->data <= head2->data) {
            moveNode(lastPtrRef, &head1);
        }
        else {
            moveNode(lastPtrRef, &head2);
        }

        lastPtrRef = &((*lastPtrRef)->next);
    }

    return mergedList;
}

void moveNode(struct Node** destRef, struct Node** sourceRef) {
    struct Node* newNode = *sourceRef;
    *sourceRef = newNode->next;
    newNode->next = *destRef;
    *destRef = newNode;
}

int main() {
    struct Node* head1 = NULL;
    struct Node* head2 = NULL;

    push(&head1, 7);
    push(&head1, 5);
    push(&head1, 3);

    push(&head2, 6);
    push(&head2, 4);
    push(&head2, 2);

    printf("List 1: ");
    printList(head1);
    printf("\nList 2: ");
    printList(head2);

    struct Node* mergedList = mergeLists(head1, head2);

    printf("\nMerged List: ");
    printList(mergedList);

    return 0;
}

希望这个代码能够帮助到你！

C语言创建两个单链表的代码

下面是C语言创建两个单链表的代码：

#include <stdio.h>
#include <stdlib.h>

struct ListNode {
    int val;
    struct ListNode *next;
};

struct ListNode* createLinkedList(int* arr, int size) {
    struct ListNode* head = NULL, *tail = NULL;
    for (int i = 0; i < size; ++i) {
        struct ListNode* node = (struct ListNode*)malloc(sizeof(struct ListNode));
        node->val = arr[i];
        node->next = NULL;
        if (!head)
            head = node;
        if (tail)
            tail->next = node;
        tail = node;
    }
    return head;
}

int main() {
    int arr1[] = { 1, 2, 3, 4, 5 };
    int arr2[] = { 6, 7, 8, 9, 10 };
    struct ListNode* l1 = createLinkedList(arr1, sizeof(arr1) / sizeof(arr1[0]));
    struct ListNode* l2 = createLinkedList(arr2, sizeof(arr2) / sizeof(arr2[0]));
    return 0;
}

该代码创建了两个单链表 `l1` 和 `l2`，分别包含了两个数组 `arr1` 和 `arr2` 中的元素。你可以根据自己的需要在代码中修改数组的内容和长度。注意，创建链表后记得释放链表所占用的内存。