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：

程式碼：

/**
 * 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 (m==n) return head;

        ListNode* left = NULL;
        ListNode* curr = head;
        for(int i = 1; i < m; ++i) {
            left = curr;
            curr = curr->next;
        }

        // reverse
 

        ListNode* end  = curr;
        ListNode* prev = curr;
        curr = curr->next;
        for(int i = m; i < n; ++i) {
            ListNode* next = curr->next;
            curr->next = prev;
            prev = curr;
            curr = next;
        }

        // link to the right-part
 

        end->next = curr;

        if (left) { left->next = prev; return head;}

        return prev;

    }
};