1. 题目描述

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

Example:

Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4

2. 解题

链表打卡题，没啥好说的，代码如下：

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

struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2) {
    struct ListNode* res = (struct ListNode *)malloc(sizeof(struct ListNode));
    res->next = NULL;
    struct ListNode* ptr = res;

    while (l1 && l2) {
        if (l1->val < l2->val) {
            ptr->next = l1;
            ptr = ptr->next;
            l1 = l1->next;
        }
        else {
            ptr->next = l2;
            ptr = ptr->next;
            l2 = l2->next;
        }
    }
    ptr->next = l1 ? l1 : l2;

    return res->next;
}

主要是想分享一下LeetCode上惊艳的递归解法：

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

struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2) {
    if (l1 == NULL) return l2;
    if (l2 == NULL) return l1;
    
    if (l1->val < l2->val) {
        l1->next = mergeTwoLists(l1->next, l2);
        return l1;
    }
    else {
        l2->next = mergeTwoLists(l1, l2->next);
        return l2;
    }
}

这个递归解法比通常解法少了很多指针的操作，最后运行时间甚至比通常解法还快，不禁感慨递归真奇妙！