【Reverse Linked List II】cpp

题目：

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.

代码：

/**
 * 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) {
        // virtual begin ListNode
        ListNode dummy(-1);
        dummy.next = head;
        // move to the m-1 ListNode
        ListNode *p = &dummy;
        for (int i = 0; i < m-1; ++i) p = p->next;
        ListNode *prev = p;
        ListNode *curr = p->next;
        for (int i = 0; i < n-m; ++i){
            ListNode *tmp = curr->next;
            curr->next = tmp->next;
            ListNode *tmp2 = prev->next;
            prev->next = tmp;
            tmp->next = tmp2;
        }
        return dummy.next;
    }
};

Tips:

这道题的思路沿用我的这一篇日志：http://www.cnblogs.com/xbf9xbf/p/4212159.html

需要考虑几种case:

m=1的情况

n=end的情况

再submit两次，就OK了。

==================================================

第二次过这道题，大体思路一下子没有完全想起来。想了一下之后，回忆起来了翻转列表类似抽书本的例子。就顺着思路把代码写出来了，一次AC。

/**
 * 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) {
            ListNode* dummpy = new ListNode(0);
            dummpy->next = head;
            // find start position
            ListNode* start = dummpy;
            for ( int i=0; i<m-1; i++ ) start = start->next;
            // reverse list between start and end
            ListNode* curr = start->next;
            for ( int i=0; i<(n-m); ++i )
            {
                ListNode* tmp = curr->next;
                curr->next = tmp->next;
                tmp->next = start->next;
                start->next = tmp;
            }
            return dummpy->next;
    }
};