题目
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
解法一
思路
两个链表合并,思路很简单,先新建一个新的链表,然后比较两个链表的元素值,把较小的链表结点放到新链表中,然后继续比较。由于两个链表的长度可能不同,最终可能出现还不为空的链表,此时直接将其接到新链表最后即可。
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 cur;
ListNode res;
if(l2 == null) return l1;
else if(l1 == null) return l2;
else{
if(l1.val <= l2.val) {
cur = l1;
l1 = l1.next;
}else{
cur = l2;
l2 = l2.next;
}
res = cur;
while(l1 != null && l2 != null) {
if(l1.val <= l2.val) {
cur.next = l1;
l1 = l1.next;
cur = cur.next;
}else{
cur.next = l2;
l2 = l2.next;
cur = cur.next;
}
}
if(l1 != null) cur.next = l1;
else cur.next = l2;
}
return res;
}
}
解法二
思路
还有一种递归的方法,如果l1的值小,那么对于l1的next和l2进行合并,然后将返回值赋给l1.next,然后返l1;反之同理。
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) {
if(l1 == null) return l2;
if(l2 == null) return l1;
if(l1.val <= l2.val) {
l1.next = mergeTwoLists(l1.next, l2);
return l1;
} else {
l2.next = mergeTwoLists(l1, l2.next);
return l2;
}
}
}