两个有序链表序列的交集 C语言

以下是C语言的代码实现：

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

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

typedef struct ListNode ListNode;

struct ListNode *getIntersectionNode(struct ListNode *head1, struct ListNode *head2) {
    ListNode* p1 = head1, *p2 = head2;
    ListNode* head = (ListNode*)malloc(sizeof(ListNode)); // 创建一个虚拟头节点
    ListNode* tail = head; // tail指向新链表的尾节点

    while (p1 && p2) {
        if (p1->val < p2->val) {
            p1 = p1->next;
        } else if (p1->val > p2->val) {
            p2 = p2->next;
        } else {
            tail->next = (ListNode*)malloc(sizeof(ListNode));
            tail->next->val = p1->val;
            tail = tail->next;
            p1 = p1->next;
            p2 = p2->next;
        }
    }

    tail->next = NULL; // 将新链表的尾节点指向NULL，表示结束
    return head->next; // 返回新链表的头节点
}

int main() {
    ListNode* head1 = (ListNode*)malloc(sizeof(ListNode));
    head1->val = 1;
    head1->next = (ListNode*)malloc(sizeof(ListNode));
    head1->next->val = 2;
    head1->next->next = (ListNode*)malloc(sizeof(ListNode));
    head1->next->next->val = 3;
    head1->next->next->next = (ListNode*)malloc(sizeof(ListNode));
    head1->next->next->next->val = 4;
    head1->next->next->next->next = NULL;

    ListNode* head2 = (ListNode*)malloc(sizeof(ListNode));
    head2->val = 2;
    head2->next = (ListNode*)malloc(sizeof(ListNode));
    head2->next->val = 4;
    head2->next->next = NULL;

    ListNode* head = getIntersectionNode(head1, head2);

    while (head) {
        printf("%d ", head->val);
        head = head->next;
    }

    return 0;
}

在这个实现中，我们同样使用了一个虚拟头节点来简化代码，使用了动态内存分配来创建节点，需要在程序结束时手动释放申请的空间。