GuilinDev

Lc0092

05 August 2008

92 - Reverse Linked List II

原题概述

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

Note: 1 ≤ m ≤ n ≤ length of list.

Example:

1
2

Input: 1->2->3->4->5->NULL, m = 2, n = 4
Output: 1->4->3->2->5->NULL

题意和分析

相对于上一道题，这道题是要求对链表从m到n的区间进行翻转（从位置1开始计算）。方法还是一样的，可以维护指针来标记遍历到哪里了，到m位置处再开始做翻转处理，然后到n位置结束翻转即可。

代码

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/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        if (head == null) {
            return null;
        }
        ListNode dummy = new ListNode(-1);
        dummy.next = head;

        ListNode pre = dummy;//记录翻转前的结点
        for (int i = 0, i < m - 1; i++) {//遍历到翻转前的结点
            pre = pre.next;
        }

        ListNode start = pre.next;//开始翻转的第一个结点
        ListNode reversing = start.next;//即将要被翻转的结点

        // 1 - 2 - 3 - 4 - 5 ; m = 2; n = 4 ---> pre = 1, start = 2, reversing = 3
        // dummy-> 1 -> 2 -> 3 -> 4 -> 5

        for (int i = 0; i < n - m; i++) {//翻转的区间
            start.next = reversing.next;
            reversing.next = pre.next;
            pre.next = reversing;
            reversing = start.next;
        }

        // 翻转1: dummy->1 - 3 - 2 - 4 - 5; pre = 1, start = 2, reversing = 4
        // 翻转2: dummy->1 - 4 - 3 - 2 - 5; pre = 1, start = 2, reversing = 5 (finish)

        return dummy.next;
    }
}
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