[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.

画图会比较容易理解

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (head == NULL)
            return NULL;
            
        ListNode *q = NULL;
        ListNode *p = head;
        for(int i = 0; i < m - 1; i++)
        {
            q = p;
            p = p->next;
        }
        
        ListNode *end = p;
        ListNode *pPre = p;
        p = p->next;
        for(int i = m + 1; i <= n; i++)
        {
            ListNode *pNext = p->next;
            
            p->next = pPre;
            pPre = p;
            p = pNext;
        }
        
        end->next = p;
        if (q)
            q->next = pPre;
        else
            head = pPre;
        
        return head;
    }
};