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 <= 1041 <= a <= b < list1.length - 11 <= 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
}
}