【LeetCode】25. Reverse Nodes in k-Group (2 solutions)

Reverse Nodes in k-Group

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

解法一：

看到逆序，第一反应就是栈。

使用栈，每k个结点进栈，再出栈，就实现了逆序。

/**
* 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* newhead = new ListNode(-1);
ListNode* tail = newhead;
ListNode* begin = head;
ListNode* end = begin;
while(true)
{
int count = k;
while(count && end != NULL)
{
end = end->next;
count --;
}
if(count == 0)
{//reverse from [begin, end)
stack<ListNode*> s;
while(begin != end)
{
s.push(begin);
begin = begin->next;
}
while(!s.empty())
{
ListNode* top = s.top();
s.pop();
tail->next = top;
tail = tail->next;
}
}
else
{//leave out
tail->next = begin;
break;
}
}
return newhead->next;
}
};

解法二：

自定义函数reverse(begin, end)

对[begin, end]范围内实现逆序，并且更新begin, end

逐k次调用即可。

注意：

(1)由于需要更新begin, end，因此参数形式为ListNode*&

(2)在[begin,end]范围内实现逆序之后，需要链如原先的链表，不可脱离

/**
* 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) {
if(head == NULL)
return NULL;
if(k == 1)
//no swap
return head;

int i = 1;
//head node
ListNode* newhead = new ListNode(-1);
newhead->next = head;
ListNode* tail = newhead;

ListNode* begin = head;
ListNode* end = begin;
while(end != NULL)
{
if(i%k == 0)
{
reverse(begin, end);
tail->next = begin;
tail = end;
//new begin
begin = end->next;
}
end = end->next;
i ++;
}
return newhead->next;
}
void reverse(ListNode*& begin, ListNode*& end)
{//reverse the list. begin points to new begin, end points to new end
if(begin == end)
{//only one node
return;
}
else if(begin->next == end)
{//two nodes
begin->next = end->next;
end->next = begin;
//swap begin and end
ListNode* temp = begin;
begin = end;
end = temp;
}
else
{//at least three nodes
ListNode* pre = begin;
ListNode* cur = pre->next;
ListNode* post = cur->next;

while(post != end->next)
{
cur->next = pre;
pre = cur;
cur = post;
post = post->next;
}
cur->next = pre;
//old begin points to the new end
end = begin;
end->next = post;
//cur points to the old end
begin = cur;
}
}
};