Problem:

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

Solution:

采用和上题基本类似的方法，只不过这次是将k个反转，每次找到一个反转对，然后采用头插法将链表反转，时间复杂度O（n）

题目大意：

给一个链表，将这个链表每K个反转。

Java源代码（330ms）：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode s,p=new ListNode(0);
        p.next=head;head=p;s=p;
        int len=k;
        while(k-->0 && s!=null)s=s.next;
        while(s!=null){
            ListNode tmp,l=p.next,flag=s.next,tail=p.next;
            p.next=null;
            while(l!=flag){
                tmp=p.next;
                p.next=l;
                l=l.next;
                p.next.next=tmp;
            }
            tail.next=l;
            p=tail;
            s=tail;
            k=len;
            while(k-->0 && s!=null)s=s.next;
        }
        return head.next;
    }
}

C语言源代码（8ms）：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
void reverse(struct ListNode** p,struct ListNode **s){
	struct ListNode *l=(*p)->next,*tmp,*tail=(*p)->next,*flag=(*s)->next;
    (*p)->next=NULL;
    while(l!=flag){
        tmp=(*p)->next;
        (*p)->next=l;
        l=l->next;
        (*p)->next->next=tmp;
    }
    tail->next=l;
    *p=tail;
    *s=tail;
}
struct ListNode* reverseKGroup(struct ListNode* head, int k) {
    struct ListNode *s,*p=(struct ListNode*)malloc(sizeof(struct ListNode));
    int len=k;
    p->next=head;
    head=p;s=p;
    while(k-- && s!=NULL)s=s->next;
    while(s!=NULL){
        reverse(&p,&s);
        k=len;
        while(k-- && s!=NULL)s=s->next;
    }
    return head->next;
}

C++源代码（30ms）：

/**
 * 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 *s,*p=(ListNode*)malloc(sizeof(ListNode));
        int len=k;
        p->next=head;head=p;s=p;
        while(k-- && s!=NULL)s=s->next;
        while(s!=NULL){
            reverse(p,s);
            k=len;
            while(k-- && s!=NULL)s=s->next;
        }
        return head->next;
    }
private:
    void reverse(ListNode* &p,ListNode* &s){
        ListNode *tmp,*tail=p->next,*flag=s->next,*l=p->next;
        p->next=NULL;
        while(l!=flag){
            tmp=p->next;
            p->next=l;
            l=l->next;
            p->next->next=tmp;
        }
        tail->next=l;
        p=tail;
        s=tail;
    }
};

Python源代码（268ms）：

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    # @param {ListNode} head
    # @param {integer} k
    # @return {ListNode}
    def reverseKGroup(self, head, k):
        p=ListNode(0)
        p.next=head;head=p;s=p
        len=k
        while k>0 and s!=None:k-=1;s=s.next
        while s!=None:
            flag=s.next;tail=p.next;l=p.next
            while l!=flag:
                tmp=p.next
                p.next=l
                l=l.next
                p.next.next=tmp
            tail.next=l
            p=tail;s=tail
            k=len
            while k>0 and s!=None:k-=1;s=s.next
        return head.next