Reverse Linked List II 解答

Question

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.

Solution

Four pointers:

Dummy node, pre, start, then, tmp

First, we find the position to start reversal.

Key point here is to use Dummy node, and pre points the (m - 1) position.

Then, we use three pointers, start, then, tmp to implement reversal. Just the same way as Reverse Linked List.

Several points to note here:

1. Move (m - 1) steps to get pre position

2. Move (n - m) steps to implement reversal

3. Link reversed sub-list with original unreversed sub-lists.

In-place and in one-pass.

/**
 * 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 (m == n || head == null)
            return head;
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode pre = dummy, start = dummy, then = dummy, tmp = dummy;
        // Move pre to (m - 1) position
        for (int i = 0; i < (m - 1); i++) {
            pre = pre.next;
        }
        // Start = m
        start = pre.next;
        then = start.next;
        start.next = null;
        // Move (n - m) steps for reverse
        for (int i = 0; i < (n - m); i++) {
            tmp = then.next;
            then.next = start;
            start = then;
            then = tmp;
        }
        pre.next.next = then;
        pre.next = start;
        return dummy.next;        
    }
}