# [LeetCode 21] Merge Two Sorted Lists

### Question

link

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.

### Stats

 Frequency 5 Diffficulty 2 Adjusted Difficulty 2 Time to use ----------

Ratings/Color = 1(white) 2(lime) 3(yellow) 4/5(red)

### Analysis

This question is easy. There are difficult ways to solve.

### Solution

There are 2 ways to solve this problem. First and the easy one is to do recursion.

Second solution, also my original solution is to use a ‘fakeHead’ to help. Read this blog.

### My code

Recursion:

``````/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) {
*         val = x;
*         next = null;
*     }
* }
*/
public class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if (l1 == null) {
return l2;
} else if (l2 == null) {
return l1;
} else {
if (l1.val < l2.val) {
l1.next = mergeTwoLists(l1.next, l2);
return l1;
} else {
l2.next = mergeTwoLists(l1, l2.next);
return l2;
}
}
}
}
``````

Temporary header + link operations

``````/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) {
*         val = x;
*         next = null;
*     }
* }
*/
public class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
ListNode pre = new ListNode(0);
ListNode cur = pre;
while (l1 != null && l2 != null) {
if (l1.val < l2.val) {
cur.next = l1;
l1 = l1.next;
} else {
cur.next = l2;
l2 = l2.next;
}
cur = cur.next;
}
if (l1 == null) cur.next = l2;
else cur.next = l1;
return pre.next;
}
}
``````