next = list1; } else { // 如果 list1 没有结点,表明 list2 已遍历完成, // 则将 list2 直接放在 tail 后面 tail.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/...
*@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;...
next pre.next = head1 if head1 else head2 return head.next2.3 解法二:递归 ## 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 ...
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
1、题目描述 2、题目分析 题目要求合并有序的两个链表,要求不能额外申请空间。 3、代码 1ListNode* mergeTwoLists(ListNode* l1, ListNode*l2) {2if( l1 == NULL )returnl2;3if( l2 == NULL )returnl1;4if( l1 == NULL && l2 == NULL )returnNULL;5ListNode* r = ( l1->val <= l2->val ...
}else if (l1.Val < l2.Val) { l1.Next = mergeTwoLists(l1.Next, l2); return l1; }else { l2.Next = mergeTwoLists(l1, l2.Next); return l2; } } 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. ...
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. AI检测代码解析 1. AI检测代码解析 Example: Input: 1->2->4, 1->3->4 Output: 1->1->2->3->4->4 ...
力扣Leetcode 21|合并两个有序链表Merge Two sorted Array 2020年10月13日 12:00595浏览·3喜欢·0评论 爱学习的饲养员 粉丝:6.7万文章:46 关注 视频讲解 622:17 Leetcode力扣 1-300题视频讲解合集|手画图解版+代码【持续更新ing】 77.6万783 视频爱学习的饲养员 ...
1. 题目描述 Merge two sorted linked lists and return it as a new sorted list. The new list ...
一、题目描述 将两个有序链表合并为一个新的有序链表并返回。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: 二、代码实现