[LeetCode]Reverse Linked List II, 解题报告 - c++编程基础

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;
    }
}