标签：code lse rip def bsp NPU div new null
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
M1: recursion
时间：O(M+N)，空间：O(M+N)
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ class Solution { public ListNode mergeTwoLists(ListNode l1, ListNode l2) { if(l1 == null) return l2; else if(l2 == null) 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; } } }
M2: iteration
时间：O(M+N)，空间：O(1)
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ class Solution { public ListNode mergeTwoLists(ListNode l1, ListNode l2) { ListNode dummy = new ListNode(-1); ListNode cur = dummy; while(l1 != null && l2 != null) { if(l1.val < l2.val) { cur.next = l1; l1 = l1.next; } else { cur.next = l2; l2 = l2.next; } cur = cur.next; } cur.next = l1 == null ? l2 : l1; return dummy.next; } }
21. Merge Two Sorted Lists - Easy
标签：code lse rip def bsp NPU div new null
原文地址：https://www.cnblogs.com/fatttcat/p/10021841.html
