LEETCODE: 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

这个和“Swap Nodes in Pairs”这题相似，唯一是复杂度稍微高一点。还是要注意指针的问题，要记住前一个Node，后一个Node等。题目其实不难！考察你对指针的理解了。

/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
bool hasKNodes(ListNode *head, int k, ListNode **newHead) {
*newHead = head;
for(int ii = 0; ii < k - 1; ii ++) {
if((*newHead) != NULL && (*newHead)->next != NULL) {
(*newHead) = (*newHead)->next;
}
else {
return false;
}
}

return true;
}

ListNode *reverseKGroup(ListNode *head, int k) {
if(k <= 1 || head == NULL) return head;
ListNode **newHead = new ListNode*();
ListNode *outputHead = NULL;
if(hasKNodes(head, k, newHead)) {
outputHead = *newHead;

ListNode* pre = NULL;
while(hasKNodes(head, k, newHead)) {
if(pre != NULL) pre->next = *newHead;
pre = head;
ListNode* curHead = head;
ListNode* after = NULL;
ListNode* before = head;
ListNode* endtag = (*newHead)->next;
for(head = head->next; head != endtag;) {
after = head->next;
head->next = before;
before = head;
head = after;
}
curHead->next = endtag;
}
}
else {
return head;
}

return outputHead;
}
};