# LeetCode 1669. Merge In Between Linked Lists

You are given two linked lists: list1 and list2 of sizes n and m respectively.

Remove list1's nodes from the ath node to the bth node, and put list2 in their place.

The blue edges and nodes in the following figure indicate the result:

Build the result list and return its head.

Example 1:

Input: list1 = [10,1,13,6,9,5], a = 3, b = 4, list2 = [1000000,1000001,1000002]
Output: [10,1,13,1000000,1000001,1000002,5]
Explanation: We remove the nodes 3 and 4 and put the entire list2 in their place. The blue edges and nodes in the above figure indicate the result.


Example 2:

Input: list1 = [0,1,2,3,4,5,6], a = 2, b = 5, list2 = [1000000,1000001,1000002,1000003,1000004]
Output: [0,1,1000000,1000001,1000002,1000003,1000004,6]
Explanation: The blue edges and nodes in the above figure indicate the result.

Constraints:

• 3 <= list1.length <= 104
• 1 <= a <= b < list1.length - 1
• 1 <= list2.length <= 104

when index hit a - 1, mark1, when hit b - 1, mark2.

mark1.next = list2.

Move list2 to the tail and point to mark2.next.

Time Compelxity; O(b + n). n = list2 length.

Space: O(1).

AC Java:

 1 /**
2  * Definition for singly-linked list.
3  * public class ListNode {
4  *     int val;
5  *     ListNode next;
6  *     ListNode() {}
7  *     ListNode(int val) { this.val = val; }
8  *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
9  * }
10  */
11 class Solution {
12     public ListNode mergeInBetween(ListNode list1, int a, int b, ListNode list2) {
13         ListNode cur = list1;
14         ListNode mark1 = null;
15         for(int i = 0; i < b; i++){
16             if(i == a - 1){
17                 mark1 = cur;
18             }
19
20             cur = cur.next;
21         }
22
23         mark1.next = list2;
24         while(list2.next != null){
25             list2 = list2.next;
26         }
27
28         list2.next = cur.next;
29         cur.next = null;
30         return list1;
31     }
32 }

posted @ 2024-05-16 11:45  Dylan_Java_NYC  阅读(2)  评论(0编辑  收藏  举报