[LeetCode]Reverse Linked List II

[LeetCode]Reverse Linked List II

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.

這道題在單鏈表的反轉上做了一點點的修改，要求只反轉鏈表的從第m個到第n個結點。分兩步走：
1. 定位到第m個結點
2. 進行反轉直到第n個結點：沒遇到一個結點，就把它插入到第m個結點的前面的位置，然後繼續下一個結點。

在操作的過程中需要注意：
1. 需要一個指向第m個結點前驅結點的指針
2. 若m==n，則無需操作。

下面貼上代碼：

/**
 * 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* first = new ListNode(0);
        int len = n - m;
        first->next = head;
        ListNode* p = first;
        while (p&&m > 1){
            p = p->next;
            m--;
        }
        ListNode* q = p->next;
        ListNode* tail = q;
        p->next = NULL;
        ListNode* r = NULL;
        while (q&&len >= 0){
            r = q->next;
            q->next = p->next;
            p->next = q;
            q = r;
            len--;
        }
        tail->next = r;
        return first->next;
    }
};