## 解法二:递归 Recursion class Solution: def mergeTwoLists(self, head1, head2): ## head1 和 head2 头结点指代两个链表 ## 递归的 base,结束条件 if head1 is None: return head2 elif head2 is None: return head1 elif head1.val < head2.val: ## 把小的值 head1 提取出来,拼接上后面...
*@returnListNode */functionmergeTwoLists($l1,$l2){if($l1==null) {return$l2; }if($l2==null) {return$l1; }if($l1->val <=$l2->val) {$l1->next =mergeTwoLists($l1->next,$l2);return$l1; }if($l2->val <=$l1->val) {$l2->next =mergeTwoLists($l2->next,$l1);return$l2;...
参考链接:https://leetcode.com/problems/merge-two-sorted-lists/discuss/9814/3-lines-c-12ms-and-c-4ms。参考代码如下: class Solution { public: ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) { if (!l1 || (l2 && l1->val > l2->val)) swap(l1, l2); if (l1) l1->next = ...
next = list2; } // 返回合并后的链表的头结点 head_pre.next } } 题目链接: Merge Two Sorted Lists : leetcode.com/problems/m 合并两个有序链表: leetcode-cn.com/problem LeetCode 日更第 52 天,感谢阅读至此的你 欢迎点赞、收藏鼓励支持小满...
Can you solve this real interview question? Merge Two Sorted Lists - You are given the heads of two sorted linked lists list1 and list2. Merge the two lists into one sorted list. The list should be made by splicing together the nodes of the first two li
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. 思路: easy 。 算法: 1. public ListNode mergeTwoLists(ListNode l1, ListNode l2) { 2. null, l, p = l1, q = l2; ...
leetcode 21. Merge Two Sorted Lists 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...
21. Merge Two Sorted Lists 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...
题目:21. Merge Two Sorted Lists 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. 翻译:合并两个排好序的链列,并将其作为新链表返回。新链表应通过将前两个列表的节点拼接在一起。
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. 题目大意:合并两个有序的链表 思路:通过比较两个链表的节点大小,采用尾插法建立链表。 代码如下: ...