Jianzhi offer record 3
☁️ Title description: Merge two ascending linked lists into a new ascending linked list and return. The new linked list is formed by splicing all the nodes of the given two linked lists.
☁️ Example 1:
输入：l1 = [1,2,4], l2 = [1,3,4] 输出：[1,1,2,3,4,4]
☁️ Example 2:
输入：l1 = [], l2 = [] 输出：[]
☁️ Example 3:
输入：l1 = [], l2 = [0] 输出：[0]
☁️ Tips:
- The range of the number of nodes of the two linked lists is
[0, 50] -100 <= Node.val <= 100l1andl2are in non-decreasing order
☁️ Method 1: Recursion
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
if(list1==null){
return list2;
}
else if(list2==null){
return list1;
}else if(list1.val<list2.val){
list1.next = mergeTwoLists(list1.next,list2);
return list1;
}else{
list2.next = mergeTwoLists(list1,list2.next);
return list2;
}
}
}
☁️ Method 2: Iteration
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
if(list1==null){
return list2;
}
else if(list2==null){
return list1;
}
ListNode head = new ListNode(-1);
ListNode cur = head;
while(list1 != null && list2 != null){
if(list1.val<list2.val){
cur.next = list1;
list1 = list1.next;
}else{
cur.next = list2;
list2 = list2.next;
}
cur =cur.next;
}
if(list1 != null){
cur.next=list1;
}else if(list2 != null){
cur.next=list2;
}
return head.next;
}
}
