“递归”: https://leetcode.com/tag/recursion/ “链表”: https://leetcode.com/tag/linked-list/ Similar Questions: “合并K个升序链表”: https://leetcode.com/problems/merge-k-sorted-lists/ “合并两个有序数组”: https://leetcode.com/problems/merge-sorted-array/ “排序链表”: https://leetcode.com/problems/sort-list/ “最短单词距离 II”: https://leetcode.com/problems/shortest-word-distance-ii/

Problem: Link to heading

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。

示例 1：

输入：l1 = [1,2,4], l2 = [1,3,4]
输出：[1,1,2,3,4,4]

示例 2：

输入：l1 = [], l2 = []
输出：[]

示例 3：

输入：l1 = [], l2 = [0]
输出：[0]

提示：

  • 两个链表的节点数目范围是 [0, 50]
  • 100 <= Node.val <= 100
  • l1 和 l2 均按 非递减顺序 排列

Solution: Link to heading

https://leetcode-cn.com/problems/merge-two-sorted-lists/solution/dong-hua-yan-shi-21-he-bing-liang-ge-you-blxu/

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        if (l1 == nullptr) {
            return l2;
        } else if (l2 == nullptr) {
            return l1;
        } else if (l1->val < l2->val) {
            l1->next = mergeTwoLists(l1->next, l2);
            return l1;
        } else {
            l2->next = mergeTwoLists(l1, l2->next);
            return l2;
        }
    }
};