Reverse Linked List II

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.

思路

这里需要操作连续的三个节点，分别是 pre  cur 和 nxt ，->口->口->口

这里 用到了4个指针，revtail 和 revhead  表示翻转后的子链表的头和尾（即m节点和n节点），newhead 和 newtail 表示翻转的起始节点和结束节点（即m-1节点和n+1节点）； 

代码一：

/**
* 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 || m>=n)
return head;
ListNode *pre = head;
ListNode *cur = head;
ListNode *nxt = cur->next;
ListNode *revtail = NULL;
ListNode *revhead = NULL;
ListNode *newtail = NULL;
ListNode *newhead = NULL;
int i = 1;
while(cur)
{
nxt = cur->next;
if(i==m)
{
revtail = cur;
newhead = pre;
pre->next = NULL;
}
if(i>m && i<n)
{
cur->next = pre;
}
if(i==n)
{
cur->next = pre;
revhead = cur;
newtail = nxt;
}
i++;
pre = cur;
cur = nxt;
}
if(newhead!=revtail)
{
newhead->next = revhead;
revtail->next = newtail;
return head;
}
else
{
revtail->next = newtail;
return revhead;
}

}
};

代码二：这里我申请了一个Dump节点作为附加的头结点。

/**
* 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 || m>=n)
return head;
ListNode dump(0);
dump.next = head;
ListNode *pre = &dump;
ListNode *cur = head;
ListNode *nxt = cur->next;
ListNode *revtail = NULL;
ListNode *revhead = NULL;
ListNode *newtail = NULL;
ListNode *newhead = NULL;
int i = 1;
while(cur && i<=n)
{
nxt = cur->next;
if(i==m)
{
revtail = cur;
newhead = pre;
}
if(i>m && i<n)
{
cur->next = pre;
}
if(i==n)
{
cur->next = pre;
revhead = cur;
newtail = nxt;
}
i++;
pre = cur;
cur = nxt;
}
newhead->next = revhead;
revtail->next = newtail;
return dump.next;
}
};

 代码三：这里减少重复的指针，尽量少的利用指针变量

/**
* 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 || m>=n)
return head;
ListNode dump(0);
dump.next = head;
ListNode *pre = &dump;
ListNode *cur = head;
ListNode *nxt = NULL;
for(int i=1;i<m;i++)
{
pre = cur;
cur = cur->next;
}
ListNode *newhead = pre;
ListNode *newtail = cur;
pre = cur;
cur = cur->next;
int j = m+1;
while(cur && j++<=n)
{
nxt = cur->next;
cur->next = pre;
pre = cur;
cur = nxt;
}
newtail->next = cur;
newhead->next = pre;
return dump.next;
}
};