Reverse Linked List II | Leetcode

Reverse a linked list from position m to n. Do it in-place and in one-pass.

For example:

Given 1->2->3->4->5->NULL, m = 2 and n = 4,
return 1->4->3->2->5->NULL.

Note:

Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.

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Solution: This is a tricky linked list question that is hard to get right the first time. The idea is quite straightforward. First, skip the first (m - 1) nodes so that you can get to one node before m. If m is the head node, then the node before head is called "beforeHead", which is an artificial node to simplify our algorithm.

Second, from the m-th node to the n-th node, reverse the list in-place.

Lastly, when the reversal is done, link up the whole list. You will need to join the node before m to the n-th node, and the m-th node to the node after n.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        if (head == null || m == n) return head;
        
        // Avoid needing to keep track of head node later on.
        ListNode beforeHead = new ListNode(0);
        beforeHead.next = head;
        
        // Skip first m-1 nodes to get to one node before m.
        ListNode prev = beforeHead;
        for (int i = m - 1; i > 0; i--) {
            prev = prev.next;
        }
        
        // Reverse from nodes m to n in-place.
        ListNode beforeMthNode = prev;
        ListNode mthNode = prev.next;
        ListNode cur = mthNode;
        for (int i = n - m; i >= 0; i--) {
            ListNode next = cur.next;
            cur.next = prev;
            prev = cur;
            cur = next;
        }
        
        // Tie up loose ends.
        beforeMthNode.next = prev; // prev is the (n-1)-th node
        mthNode.next = cur;        // cur is the n-th node
        
        return beforeHead.next;
    }
}