合并两个已排序的链接列表并将其作为新列表返回。新列表应该通过拼接前两个列表的节点来完成。
例:
输入: 1-> 2-> 4,1-> 3-> 4
输出: 1-> 1-> 2-> 3-> 4-> 4
直接递归实现,代码来了
/**
* 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;
}
if(l2==null) {
return l1;
}
while(l1!=null||l2!=null){
if(l1.val<l2.val){
l1.next=mergeTwoLists(l1.next,l2);
return l1;
}else{
l2.next=mergeTwoLists(l1,l2.next);
return l2;
}
}
return null;
}
}
非递归也不能少,比递归运行时间短(粘贴大神代码)
/**
* 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 head = new ListNode(0);
ListNode current = head;
while(l1 != null && l2 !=null) {
int d1 = l1.val;
int d2 = l2.val;
if (d1 <= d2) {
current.next = new ListNode(d1);
current = current.next;
l1=l1.next;
}
else
{current.next = new ListNode(d2);
current = current.next;
l2=l2.next;}
}
while(l1 == null && l2 != null) {
current.next = new ListNode(l2.val);
current = current.next;
l2=l2.next;}
while(l2 == null && l1 !=null) {
current.next = new ListNode(l1.val);
current = current.next;
l1=l1.next;}
return head.next;
}
}