翻转链表 II - LintCode

描述
翻转链表中第m个节点到第n个节点的部分

m，n满足1 ≤ m ≤ n ≤ 链表长度

样例
给出链表1->2->3->4->5->null， m = 2 和n = 4，返回1->4->3->2->5->null

挑战
在原地一次翻转完成

思路

#ifndef C036_H
#define C036_H
#include<iostream>
using namespace std;
class ListNode{
public:
    int val;
    ListNode *next;
    ListNode(int val){
        this->val = val;
        this->next = next;
    }
};
class Solution {
public:
    /**
    * @param head: ListNode head is the head of the linked list
    * @param m: An integer
    * @param n: An integer
    * @return: The head of the reversed ListNode
    */
    ListNode * reverseBetween(ListNode * head, int m, int n) {
        // write your code here
        //当m等于n,不翻转
        if (m == n)
            return head;
        ListNode *pn = head, *pm = head;
        ListNode *node = new ListNode(-1);
        node->next = head;
        ListNode *pre = node;
        //找到第m个节点和其前一个节点
        for (int i = 1; i < m; ++i)
        {
            pm = pm->next;
            pre = pre->next;
        }
        //找到第n个节点
        for (int i = 1; i < n; ++i)
        {
            pn = pn->next;
        }
        ListNode *post = pn->next;//post表示第m个节点的后一个节点
        ListNode *start = pre;//start表示第n个节点的前一个节点
        //pre表示要翻转的第一个节点,pm表示要翻转的第二个节点
        pre = pre->next;
        pm = pm->next;
        //保存当前节点pm,前一个节点pre以及后一个节点pnext
        //令pm->next=pre再处理各节点
        for (int i = 0; i < n - m; ++i)
        {
            ListNode *pnext = pm->next;
            pm->next = pre;
            if (i == 0)
                pre->next = post;
            pre = pm;
            pm = pnext;
        }
        start->next = pre;
        return node->next;
    }
};
#endif