LeetCode 1669. Merge In Between Linked Lists

原题链接在这里：https://leetcode.com/problems/merge-in-between-linked-lists/description/

题目：

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

题解：

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:

/**
 * 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 mergeInBetween(ListNode list1, int a, int b, ListNode list2) {
        ListNode cur = list1;
        ListNode mark1 = null;
        for(int i = 0; i < b; i++){
            if(i == a - 1){
                mark1 = cur;
            }

            cur = cur.next;
        }

        mark1.next = list2;
        while(list2.next != null){
            list2 = list2.next;
        }

        list2.next = cur.next;
        cur.next = null;
        return list1;
    }
}