LeetCode 92. Reverse Linked List II（翻转链表）

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本文介绍了一种在原地且仅一次遍历的情况下反转链表中从位置 m 到 n 的节点的方法。通过计数器来定位需要反转的部分，并实现反转操作。

原题网址：https://leetcode.com/problems/reverse-linked-list-ii/

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.

方法：计数器。

/**
 * 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 || head.next == null || m == n) return head;
        int dist = n-m;
        if (dist == 0) return head;
        ListNode start = new ListNode(0);
        start.next = head;
        ListNode reverseTail = null;
        ListNode reversePrev = null;
        ListNode prev = start;
        ListNode current = start.next;
        for(int i=1; i<=n; i++) {
            ListNode next = current.next;
            if (i==m) {
                reversePrev = prev;
                reverseTail = current;
            } else if (i==n) {
                next = current.next;
                current.next = prev;
                reversePrev.next = current;
                reverseTail.next = next;
            } else if (m<i && i<n) {
                next = current.next;
                current.next = prev;
            }
            prev = current;
            current = next;
        }
        return start.next;
    }
}