Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.
If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.
You may not alter the values in the nodes, only nodes itself may be changed.
Only constant memory is allowed.
For example,
Given this linked list: 1->2->3->4->5
For k = 2, you should return: 2->1->4->3->5
For k = 3, you should return: 3->2->1->4->5
每次选取好一个group后,调用翻转链表子函数翻转该group
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/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class
Solution {
public
:
ListNode *reverseKGroup(ListNode *head,
int
k) {
ListNode tmphead(0);
ListNode *p = head, *tail = &tmphead;
//tail 为当前已经处理完的尾节点
while
(p)
{
ListNode *groupHead = p;
int
i = 1;
for
(i = 1; i < k && p->next != NULL; i++)p = p->next;
if
(i != k){tail->next = groupHead;
return
tmphead.next;}
//最后不足k个节点
p = p->next;
pair<ListNode*, ListNode*>ht = reverseList(groupHead, p);
tail->next = ht.first;
tail = ht.second;
}
return
tmphead.next;
}
//翻转链表,并返回翻转后链表的头结点和尾节点
pair<ListNode*, ListNode*> reverseList(ListNode* head, ListNode *end)
{
ListNode *p = head, *newHead = NULL;
while
(p != end)
{
ListNode *tmp = p->next;
p->next = newHead;
newHead = p;
p = tmp;
}
return
make_pair(newHead, head);
}
};
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上面的方法实际上遍历了2遍链表,可以稍加改进在计数k个链表的过程中就开始翻转, 最后部分如果不足k个节点,则还原原来的顺序 本文地址
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/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class
Solution {
public
:
ListNode *reverseKGroup(ListNode *head,
int
k) {
ListNode tmphead(0);
ListNode *p = head, *tail = &tmphead;
//tail 为当前已经处理完的尾节点
while
(p)
{
pair<ListNode*, ListNode*>ht = reverseList(p,k);
tail->next = ht.first;
tail = ht.second;
p = tail->next;
}
return
tmphead.next;
}
//翻转从head开始的k个链表节点，如果不足k个则不翻转，返回翻转后的这k个节点
//子链表的head和tail。翻转后的子链表还是和子链表后面的部分相连的。
pair<ListNode*, ListNode*> reverseList(ListNode* head,
int
k)
{
ListNode *p = head, *newHead = NULL;
int
cnt = 0;
while
(p && cnt < k)
{
ListNode *tmp = p->next;
p->next = newHead;
newHead = p;
p = tmp;
cnt++;
}
head->next = p;
//把翻转后的子链表和子链表后面的部分连起来
if
(cnt == k)
return
make_pair(newHead, head);
else
return
reverseList(newHead, cnt);
//不足k个，还原原来的顺序
}
};
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本文转自tenos博客园博客，原文链接：http://www.cnblogs.com/TenosDoIt/p/3794505.html，如需转载请自行联系原作者
