LeetCode in Kotlin

1669. Merge In Between Linked Lists

Medium

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 = [0,1,2,3,4,5], a = 3, b = 4, list2 = [1000000,1000001,1000002]

Output: [0,1,2,1000000,1000001,1000002,5]

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

Solution

import com_github_leetcode.ListNode

/*
 * Example:
 * var li = ListNode(5)
 * var v = li.`val`
 * Definition for singly-linked list.
 * class ListNode(var `val`: Int) {
 *     var next: ListNode? = null
 * }
 */
@Suppress("NAME_SHADOWING")
class Solution {
    fun mergeInBetween(list1: ListNode?, a: Int, b: Int, list2: ListNode?): ListNode? {
        var list2 = list2
        var start = list1
        for (i in 1 until a) {
            start = start!!.next
        }
        var end = start
        for (i in a..b) {
            end = end!!.next
        }
        start!!.next = list2
        while (list2!!.next != null) {
            list2 = list2.next
        }
        list2.next = end!!.next
        return list1
    }
}

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