Leetcode.com/problems/re…

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Given the head of a singly linked list, reverse the list, and return the reversed list.

Example 1:

Input: head = [1,2,3,4,5]
Output: [5,4,3,2,1]
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Example 2:

Input: head = [1,2]
Output: [2,1]
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Example 3:

Input: head = []
Output: []
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Constraints:

The number of nodes in the list is the range [0, 5000].

-5000 <= Node.val <= 5000

Follow up: A linked list can be reversed either iteratively or recursively. Could you implement both?

Solution a:

Iterative solution.

/** * Example: * var li = ListNode(5) * var v = li.`val` * Definition for singly-linked list. * class ListNode(var `val`: Int) { * var next: ListNode? = null * } */ class Solution { fun reverseList(head: ListNode?) : ListNode? { if (head? .next == null) { return head } var cur: ListNode? = head var pre: ListNode? = null while (cur ! = null) { val tempNode = cur.next cur.next = pre pre = cur cur = tempNode } return pre } }Copy the code

Method 2:

Recursively.

/** * Example: * var li = ListNode(5) * var v = li.`val` * Definition for singly-linked list. * class ListNode(var `val`: Int) { * var next: ListNode? = null * } */ class Solution { fun reverseList(head: ListNode?) : ListNode? { if (head? .next == null) { return head } val listNode = reverseList(head.next) head.next? .next = head head.next = null return listNode } }Copy the code