Reverse Linked List II

原題：
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.

解題：
這道題其實就是鏈表逆序的變種，相當於抽取了鏈表的一部分進行逆序，然後再拼接起來。但是拼接的過程中有一些異常情況要處理，比如m恰好等1或者n恰好爲鏈表的長度。其他的就很簡單了，可以AC的C++代碼如下：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:

ListNode *reverse(ListNode *head){
    if(!head || !head->next){
        return head;
    }

    ListNode *first = head;
    ListNode *second = head->next;
    ListNode *temp = NULL;
    while(second){
        temp = second->next;
        second->next = first;
        first = second;
        second = temp;
    }
    head->next = NULL;

    return first;
}

ListNode *reverseBetween(ListNode *head, int m, int n) {
    if(m > n || m < 1 || n < 1)
         return NULL;
    if(m == n){
        return head;
    }
    int length = 0;
    ListNode *pHead = head;
    while(pHead){
        length ++;
        pHead = pHead->next;
    }
    pHead = head;
    if(length < n){
        return NULL;
    }

    ListNode *start, *end, *middle = NULL;
    length = 0;
    while(pHead){
        length ++;
        if(length == m)
            start = pHead;
        if(length == m-1)
            middle = pHead;

        if(length == n)
            end = pHead;
        pHead = pHead->next;
    }

    pHead = end->next;
    end->next = NULL;
    ListNode *part = reverse(start);
    if(middle)
        middle->next = part;
    else
        head = part;
    while(part && part->next){
        part = part->next;
    }
    part->next = pHead;

    return head;
}
};