又是一道链表的题，关于链表的题好像都是关于指针的，题目都不难吧，主要是需要细心，因为不难，所以代码没有自己写，这里还可以考虑之前做过的链表翻转m至n那道题，然后调用函数就解决了······················
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* reverse(ListNode* head){
ListNode* prev = nullptr;
ListNode* nowNode = head;
while(nowNode){
ListNode* next = nowNode -> next;
nowNode -> next = prev;
prev = nowNode;
nowNode = next;
}
return prev;
}
ListNode *reverseKGroup(ListNode *head, int k) {
if(head == nullptr || head -> next == nullptr || k < 2) return head;
int cnt = 1;
ListNode* nowHead = head;
ListNode* nowNode = head;
while(nowNode && cnt < k){
cnt ++;
nowNode = nowNode -> next;
}
if(nowNode && cnt == k){
ListNode* tail = reverseKGroup(nowNode -> next , k);
nowNode -> next = nullptr;
head = reverse(head);
nowHead -> next = tail;
}
return head;
}
};
原文：http://www.cnblogs.com/qiaozhoulin/p/4775903.html