【LeetCode】Reverse Linked List II 解题报告

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【题目】

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.

【解析】

题意：把链表中m到n的部分反转（1<=m<=n<=length）。

注意要求：在原地反转，也就是不能申请额外的空间，且只能遍历一遍。

自然而然想到把链表分为三部分，重点是如何遍历一遍把中间部分反转？借助两个指针，tmpHead和tmpNext，tmpHead是遍历反转后的head，tmpNext始终是tmpHead反转后的next。

直接看代码吧，代码中有注释，不懂的请留言。

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } */public class Solution {    // Divide the list to three parts:     // 1)Nodes before m keep still;     // 2)Traverse m~n nodes;    // 3)Nodes after n keep still.    public ListNode reverseBetween(ListNode head, int m, int n) {        if (m == n) return head;                ListNode preHead = new ListNode(0);        preHead.next = head;                // The (m-1) node is the tail of first tail.        ListNode firstTail = preHead;        int k = m - 1;        while (k-- > 0) {            firstTail = firstTail.next;        }                // The m-th node is the traversed tail.        ListNode secondTail = firstTail.next;                // Traverse m~n nodes, and get the traversed head.        ListNode tmpHead = null;        ListNode tmpNext = null;        ListNode node = firstTail.next;        k = n - m + 1;        while (k-- > 0) {            tmpHead = node;            node = node.next;                        tmpHead.next = tmpNext;            tmpNext = tmpHead;        }                // Connect three parts.        firstTail.next = tmpHead;        secondTail.next = node;                return preHead.next;    }}

指针关系比较复杂，想清楚了在写代码。

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