【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：/article/5261238.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;
}
};