方法一:递归
1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode() {} 7 * ListNode(int val) { this.val = val; } 8 * ListNode(int val, ListNode next) { this.val = val; this.next = next; } 9 * } 10 */ 11 class Solution { 12 public ListNode mergeTwoLists(ListNode l1, ListNode l2) { 13 if(l1 == null) 14 return l2; 15 else if(l2 == null) 16 return l1; 17 else if(l1.val <= l2.val){ 18 l1.next = mergeTwoLists(l1.next,l2); 19 return l1; 20 }else{ 21 l2.next = mergeTwoLists(l1,l2.next); 22 return l2; 23 } 24 } 25 }
方法二:迭代法
1 class Solution { 2 public ListNode mergeTwoLists(ListNode l1, ListNode l2) { 3 //新建一个头结点用来存放结果 4 ListNode head = new ListNode(0); 5 ListNode res = head; 6 while(l1 != null && l2 != null){ 7 if(l1.val <= l2.val){ 8 res.next = l1; 9 l1 = l1.next; 10 //System.out.println(res.val); 11 }else{ 12 res.next = l2; 13 l2 = l2.next; 14 } 15 res = res.next; 16 } 17 res.next = l1 == null ? l2 : l1; 18 return head.next; 19 } 20 }