[LeetCode]Reverse Linked List II, 解题报告

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.

思路

挺不错的问题，写代码之前一定要理清自己的思路，最好能手写一下伪代码，我感觉这样会事半功倍（虽然之前我也很讨厌这样做）

我按照四个步骤来处理这道题目：

处理特殊的情况
找到m前一个节点pre
从m节点开始向后reverse，一直到n节点，并记录n之后的post节点
链接pre，post，和m～n之前的翻转链表

AC代码

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

// 1. add safe node
ListNode safeNode = new ListNode(Integer.MIN_VALUE);
safeNode.next = head;

// 2. find previous node and first switch node
ListNode pre, swOne;
pre = safeNode;
swOne = pre.next;

for (int i = 1; i < m; i++) {
pre = swOne;
swOne = swOne.next;
}

// 3. find last switch node and reverse first to last and find post node
ListNode tmpPre, tmpCur, post;
tmpPre = swOne;
tmpCur = swOne.next;
post = swOne.next;

for (int i = m; i < n; i++) {
post = tmpCur.next;
tmpCur.next = tmpPre;
tmpPre = tmpCur;
tmpCur = post;
}

// 4. link previous node, switch linked list, post node
pre.next = tmpPre;
swOne.next = post;

return safeNode.next;
}
}