[LeetCode]92.Reverse Linked List II

【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.

【题意】

将给定链表第m个节点到第n个节点的位置逆序，返回逆序后的链表。
给定M，N满足以下条件： 
1≤M≤N≤列表的长度。

【分析】

思路1：

前m-1个不变，从第m+1个到第n个，依次删除，用尾插法插入到第m-1个节点后面。

    第一步把4节点删除放入2节点之后

 第二步把5节点删除放入2节点之后

【代码1】

/**********************************   日期：2014-01-28*   作者：SJF0115*   题号: Reverse Linked List II*   来源：http://oj.leetcode.com/problems/reverse-linked-list-ii/*   结果：AC*   来源：LeetCode*   总结：**********************************/#include <iostream>#include <stdio.h>#include <algorithm>using namespace std;struct ListNode {    int val;    ListNode *next;    ListNode(int x) : val(x), next(NULL) {}};class Solution {public:    ListNode *reverseBetween(ListNode *head, int m, int n) {        if(m > n || n < 0){            return head;        }        ListNode *tail,*p,*rTail,*pre = NULL;        //添加虚拟头结点(便于反转全部)        ListNode *beginNode = (ListNode*)malloc(sizeof(ListNode));        beginNode->next = head;        pre = beginNode;        int index = 1;        //遍历前m-1个节点        while(pre != NULL && index < m){            pre = pre->next;            index++;        }        tail = pre;        rTail = pre->next;        index = 1;        //删除第m+1节点开始        while(index < (n-m+1) ){            //删除p节点            p = rTail->next;            rTail->next = p->next;            //尾插法            p->next = tail->next;            tail->next = p;            index++;        }        return beginNode->next;    }};int main() {    Solution solution;    int A[] = {1,2,3,4,5,6,7,8,9};    ListNode *head = (ListNode*)malloc(sizeof(ListNode));    head->next = NULL;    ListNode *node;    ListNode *pre = head;    for(int i = 0;i < 1;i++){        node = (ListNode*)malloc(sizeof(ListNode));        node->val = A[i];        node->next = NULL;        pre->next = node;        pre = node;    }    head = solution.reverseBetween(head->next,1,8);    while(head != NULL){        printf("%d ",head->val);        head = head->next;    }    return 0;}

【代码2】

class Solution {public:    ListNode *reverseBetween(ListNode *head, int m, int n) {        ListNode dummy(0);        dummy.next = head;        ListNode *preM, *pre = &dummy;        for (int i = 1; i <= n; ++i) {            //preM 第m-1个节点            if (i == m) preM = pre;            if (i > m && i <= n) {                //删除head节点                pre->next = head->next;                head->next = preM->next;                //尾插法                preM->next = head;                head = pre; // head has been moved, so pre becomes current            }            pre = head;            head = head->next;        }        return dummy.next;    }};