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【LeetCode】3. Merge two sorted lists · merge two ordered linked lists

2022-07-03 07:42:00 【AQin1012】

Title Description

Description in English

You are given the heads of two sorted linked lists list1 and list2. Merge the two lists in a one sorted list. The list should be made by splicing together the nodes of the first two lists. Return the head of the merged linked list.

English address

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

Description of Chinese version

Merge two ascending linked lists into a new Ascending Link 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]

Constraints· 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

Address of Chinese version

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

Their thinking

My first reaction was to read all , Then compare one by one , Then put it into a new linked list , Then I thought about it , The linked lists of these two inputs are already in order , In fact, you can read data from two linked lists and compare them , The smaller one is stored in the new linked list and points to the next data of the linked list , Big continues to compare with this new data …… repeat , Until a linked list is empty , Add the remaining data of the non empty linked list to the end of the new linked list , Just return to the new linked list .

How to solve the problem

My version

Recursive method

  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;
        }
    }
  
  //    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;
        }
    }

Official edition

public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode prehead = new ListNode(-1);

        ListNode prev = prehead;
        while (l1 != null && l2 != null) {
            if (l1.val <= l2.val) {
                prev.next = l1;
                l1 = l1.next;
            } else {
                prev.next = l2;
                l2 = l2.next;
            }
            prev = prev.next;
        }

        //  After the merger  l1  and  l2  At most, only one has not been merged yet , We can directly point the end of the linked list to the list that has not been merged 
        prev.next = l1 == null ? l2 : l1;

        return prehead.next;
    }
    
 //    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;
        }
    }