.将两个升序链表合并为一个新的升序链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。C语言回复

可以使用指针来遍历两个链表，比较两个链表的当前节点大小，将较小的节点接入新链表中，直到一个链表遍历完毕，再将另一个链表中剩余的节点接入新链表。以下是示例代码：

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

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

struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2){
    struct ListNode *newList = (struct ListNode*)malloc(sizeof(struct ListNode));
    newList->next = NULL;
    struct ListNode *cur = newList;
    while (l1 && l2) {
        if (l1->val <= l2->val) {
            cur->next = l1;
            l1 = l1->next;
        } else {
            cur->next = l2;
            l2 = l2->next;
        }
        cur = cur->next;
    }
    cur->next = l1 ? l1 : l2;
    return newList->next;
}

int main() {
    struct ListNode *l1 = (struct ListNode*)malloc(sizeof(struct ListNode));
    l1->val = 1;
    l1->next = (struct ListNode*)malloc(sizeof(struct ListNode));
    l1->next->val = 2;
    l1->next->next = (struct ListNode*)malloc(sizeof(struct ListNode));
    l1->next->next->val = 4;
    l1->next->next->next = NULL;
    struct ListNode *l2 = (struct ListNode*)malloc(sizeof(struct ListNode));
    l2->val = 1;
    l2->next = (struct ListNode*)malloc(sizeof(struct ListNode));
    l2->next->val = 3;
    l2->next->next = (struct ListNode*)malloc(sizeof(struct ListNode));
    l2->next->next->val = 4;
    l2->next->next->next = NULL;
    struct ListNode *newList = mergeTwoLists(l1, l2);
    while (newList) {
        printf("%d ", newList->val);
        newList = newList->next;
    }
    return 0;
}

输出结果为：1 1 2 3 4 4