【LeetCode】21. Merge two ordered linked lists - go language solution

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One 、 Title Description

Merge two ascending lists into a new ascending list and return . The new linked list is made up of all the nodes of the given two linked lists .

Example 1：

 Input ：l1 = [1,2,4], l2 = [1,3,4]
 Output ：[1,1,2,3,4,4]

Example 2：

 Input ：l1 = [], l2 = []
 Output ：[]

Example 3：

 Input ：l1 = [], l2 = [0]
 Output ：[0]

Tips ：

 The range of the number of nodes in the two linked lists is  [0, 50]
-100 <= Node.val <= 100
l1  and  l2  All according to   Non decreasing order   array 

Original link ：

https://leetcode.cn/problems/merge-two-sorted-lists/

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Two 、 My explanation

Their thinking

1. Create a third new linked list head , its head First, point to the linked list with a small head , And then constantly update the new linked list Next Domain to complete the merge ;
2. Tail of new linked list tail Initialize to its head ;
3. Keep comparing list1 and list2 Current Val , The smaller ones are used as new tables tail Of Next ;
4. until list1 and list2 One of the tables is over , Just add the rest of the other table to the new table tail Just go ;
5. return head .

Go The language code

/** * Definition for singly-linked list. * type ListNode struct { * Val int * Next *ListNode * } */
func mergeTwoLists(list1 *ListNode, list2 *ListNode) *ListNode {
    
    if list1 == nil && list2 == nil{
    
        return nil
    }
    if list1 == nil{
    
        return list2
    }
    if list2 == nil{
    
        return list1
    }
    
    var head,tail *ListNode
    //  Small change 
    if list1.Val <= list2.Val{
    
       head = list1
       list1 = list1.Next
    }else{
    
       head = list2
       list2 = list2.Next
    }
    
    tail = head
    for list1 != nil && list2 != nil{
    
        if list1.Val <= list2.Val{
    
            tail.Next = list1
            tail = tail.Next
            list1 = list1.Next
        }else{
    
            tail.Next = list2
            tail = tail.Next
            list2 = list2.Next
        }
    }

    if list1 == nil && list2 == nil{
    
        return head
    }else if list1 == nil{
    
        tail.Next = list2
        return head
    }else{
    
        tail.Next = list1
        return head
    }

}

Judge the result ：