leetcode 21.Merge Two Sorted Lists ,java】的更多相关文章

题目: 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. 解决方案:276ms /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next…
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 和88. Merge Sorted Array类似,数据…
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 这道混合插入有序链表和我之前那篇混合插入有序数组非常的相…
合并链表 Runtime: 4 ms, faster than 100.00% of C++ online submissions for Merge Two Sorted Lists. class Solution { public: ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) { //1 2 4 . 1 3 4 ListNode *res = ); ListNode *cur = res; while (l1 != NULL &&am…
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. 解题思路: 题目的意思是将两个有序链表合成一个有序链表. 逐个比较加入到新的链表即可. 代码如下: public static ListNode mergeTwoLists(ListNode l1, 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. 题目标签:Linked List 题目给了我们两个lists,让我们有序的合并两个 lists. 这题利用递归可以从list 的最后开始向前链接nodes,代码很简洁,清楚. Java Solution: Runti…
题目 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 合并两个有序的链表 Hints Related Topics: LinkedList 参考 归并排序-维基百科,可以递归也可以迭代,基本的链表操作 代码 Java /** * Definition for…
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. public ListNode mergeTwoLists(ListNode l1, ListNode l2) { ListNode p1 = l1; ListNode p2 = l2; ListNode fakeH…
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. 问题:将两个已排序的列表,合并为一个有序列表. 令 head 为两个列表表头中较小的一个,令 p 为新的已排序的最后一个元素.令 l1, l2 分别为两个列表中未排序部分的首节点.依次将 l1, l2 中的较小值追加…
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. 递归实现: /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(i…