LeetCode in Kotlin
92. Reverse Linked List II
Medium
Given the head of a singly linked list and two integers left and right where left <= right, reverse the nodes of the list from position left to position right, and return the reversed list.
Example 1:
Input: head = [1,2,3,4,5], left = 2, right = 4
Output: [1,4,3,2,5]
Example 2:
Input: head = [5], left = 1, right = 1
Output: [5]
Constraints:
- The number of nodes in the list is
n. 1 <= n <= 500-500 <= Node.val <= 5001 <= left <= right <= n
Follow up: Could you do it in one pass?
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 reverseBetween(head: ListNode?, left: Int, right: Int): ListNode? {
var head = head
var right = right
if (left == right) {
return head
}
var prev: ListNode? = null
var temp = head
val start: ListNode?
var k = left
while (temp != null && k > 1) {
prev = temp
temp = temp.next
k--
}
if (left > 1 && prev != null) {
prev.next = null
}
var prev1: ListNode? = null
start = temp
while (temp != null && right - left >= 0) {
prev1 = temp
temp = temp.next
right--
}
if (prev1 != null) {
prev1.next = null
}
if (left > 1 && prev != null) {
prev.next = reverse(start)
} else {
head = reverse(start)
prev = head
}
while (prev!!.next != null) {
prev = prev.next
}
prev.next = temp
return head
}
fun reverse(head: ListNode?): ListNode? {
var p: ListNode?
var q: ListNode?
var r: ListNode? = null
p = head
while (p != null) {
q = p.next
p.next = r
r = p
p = q
}
return r
}
}