c语言有序链表的交集

要实现有序链表的交集，可以先遍历两个链表，将它们的共同元素放入一个新的链表中。具体实现可以使用双指针，分别指向两个链表的头节点，比较节点的值，如果相等则将该节点加入新链表中，否则将值较小的节点向后移动一位。直到其中一个链表遍历完毕。

以下是示例代码：

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

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

struct ListNode* createList(int* nums, int numsSize) {
    struct ListNode* head = NULL;
    struct ListNode* tail = NULL;
    for (int i = 0; i < numsSize; i++) {
        struct ListNode* node = (struct ListNode*)malloc(sizeof(struct ListNode));
        node->val = nums[i];
        node->next = NULL;
        if (head == NULL) {
            head = node;
            tail = node;
        } else {
            tail->next = node;
            tail = node;
        }
    }
    return head;
}

struct ListNode* getIntersectionNode(struct ListNode* headA, struct ListNode* headB) {
    struct ListNode* dummy = (struct ListNode*)malloc(sizeof(struct ListNode));
    dummy->next = NULL;
    struct ListNode* tail = dummy;
    while (headA != NULL && headB != NULL) {
        if (headA->val == headB->val) {
            struct ListNode* node = (struct ListNode*)malloc(sizeof(struct ListNode));
            node->val = headA->val;
            node->next = NULL;
            tail->next = node;
            tail = node;
            headA = headA->next;
            headB = headB->next;
        } else if (headA->val < headB->val) {
            headA = headA->next;
        } else {
            headB = headB->next;
        }
    }
    return dummy->next;
}

void printList(struct ListNode* head) {
    while (head != NULL) {
        printf("%d ", head->val);
        head = head->next;
    }
    printf("\n");
}

int main() {
    int nums1[] = {1, 2, 3, 4, 5};
    int nums2[] = {2, 4, 6};
    struct ListNode* head1 = createList(nums1, 5);
    struct ListNode* head2 = createList(nums2, 3);
    struct ListNode* intersection = getIntersectionNode(head1, head2);
    printList(intersection); // 输出 2 4
    return 0;
}