Leetcode Reverse Linked List II

Reverse a linked list from position m ton. 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.

看到题目的第一想法，就是经典的链表逆序的变形，思想简单。

代码如下：

/** * Definition for singly-linked list. * 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(n==m)            return head;        ListNode* curr = new ListNode(0);        ListNode* pre = curr;        curr->next = head;        int count = 1;        while(count < m)        {            curr = curr->next;            count++;        }        ListNode* thead = curr;        ListNode* temp;        curr = curr->next->next;        count++;        while(count <= n)        {            temp = curr->next;            curr->next = thead->next;            thead->next = curr;            curr = temp;            count++;        }        count=0;        while(count < n-m+1)        {            thead = thead->next;            count++;        }        if(thead != NULL)            thead->next = curr;        return pre->next;    }};

代码有点混乱，进了整理后如下：

/** * Definition for singly-linked list. * 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(n==m)            return head;        n -= m;        ListNode* thead = new ListNode(0);        ListNode* pre = thead;        thead->next = head;        while(--m)            thead = thead->next;          ListNode* temp;        ListNode* pstart = thead->next;        while(n--)        {            temp = pstart->next;            pstart->next = temp->next;            temp -> next = thead->next;            thead->next = temp;        }        return pre->next;    }};