标签:链表 lse ret tno public bsp class 通过 lin
/*
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:
输入:1->2->4, 1->3->4
输出:1->1->2->3->4->4
Definition for singly-linked list.
public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
*/
1 class Solution21 { 2 3 4 /* 5 思路:迭代 时间O(n) 6 */ 7 public ListNode mergeTwoLists(ListNode l1, ListNode l2) { 8 if (l1 == null) { 9 return l2; 10 } 11 if (l2 == null) { 12 return l1; 13 } 14 15 ListNode newList = new ListNode(0);//headNode 16 ListNode nodePos = newList; 17 18 while (l1 != null && l2 != null) { 19 /* 20 将list1中结点放入newList 21 */ 22 if (l1.val <= l2.val) { 23 nodePos.next = l1; 24 l1 = l1.next; 25 } else { 26 /* 27 将list2中结点放入newList 28 */ 29 nodePos.next = l2; 30 l2 = l2.next; 31 } 32 nodePos = nodePos.next; 33 } 34 nodePos.next = l1 == null ? l2 : l1; 35 return newList.next; 36 } 37 38 /* 39 思路:递归 时间O(n) 40 */ 41 public ListNode mergeTwoLists1(ListNode l1, ListNode l2) { 42 if (l1 == null) { 43 return l2; 44 } 45 if (l2 == null) { 46 return l1; 47 } 48 if (l1.val >= l2.val) { 49 l1.next = mergeTwoLists(l1.next, l2); 50 return l1; 51 } else { 52 l2.next = mergeTwoLists(l1, l2.next); 53 return l2; 54 } 55 } 56 }
标签:链表 lse ret tno public bsp class 通过 lin
原文地址:https://www.cnblogs.com/rainbow-/p/10269262.html