Related Topics: Link to heading
“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
/**
* 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;
}
}
};