[leetcode]Reverse Nodes in k-Group

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

解题思路：双指针先用一个指针找到第k-1个位置，第一个指针指向开始位置，然后将这段元素反转，记住第k-1个元素的位置与其后
一个的位置。一次进行遍历所有元素。

/**
* 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 || k <= 1) return head;
int tk = k - 1;
ListNode *nhead = head,*ptr = head, *curPtr = head;
ListNode *endPtr = NULL;
ListNode  *tPtr1 = NULL, *tPtr2 = NULL, *prePtr = NULL, *lastPtr = NULL;while(ptr){
endPtr = ptr->next; //内层的结束标志
lastPtr = ptr->next; //翻转后的后续结点
if(tk == 0){
tPtr1 = curPtr;
while(curPtr != endPtr){
tPtr2 = curPtr;
curPtr = curPtr->next;
if(prePtr == NULL){
tPtr2->next = lastPtr;
nhead = tPtr2;
}else{
tPtr2->next = lastPtr;
prePtr->next = tPtr2;
}
lastPtr = tPtr2;
}
tk = k - 1;
prePtr = tPtr1;
ptr = endPtr;
curPtr = endPtr;
}else{
tk--;
ptr = ptr->next;
}
}
return nhead;
}
};