题目
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
思路
利用递归,将两个链表值比较后插入当前链表的下一位,依次不断递归得到合并链表
/**
* 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 result = null;
if(l1==null){
return l2;
}
if(l2==null){
return l1;
}
if(l1.val>l2.val){
result=l2;
result.next=mergeTwoLists(l1, l2.next);
}else{
result=l1;
result.next=mergeTwoLists(l1.next, l2);
}
return result;
}
}
优化思路
因为递归是不断的调用函数自身,则在调用时需要地址保存和参数传递等操作,这种额外的工作,势必会影响效率,则可考虑使用循环来解决。
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if(l1==null)return l2;
if(l2==null)return l1;
ListNode head = null, index = null, current1 = l1, current2 = l2;
if(current1.val<=current2.val){
head = index= current1;
current1=current1.next;
}else{
head = index = current2;
current2 = current2.next;
}
while(current1!=null&¤t2!=null){
if(current1.val<=current2.val){
index = index.next = current1;
current1 = current1.next;
}else{
index = index.next = current2;
current2 = current2.next;
}
}
while(current1!=null){
index = index.next = current1;
current1 = current1.next;
}
while(current2!=null){
index = index.next = current2;
current2 = current2.next;
}
return head;
}