25. 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.

k is a positive integer and is less than or equal to the length of the linked 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

题目：把每k个节点反转。

先看怎么来把每k个节点反过来：

每次把nex指的节点放在pre后面，cur之后nex之前的子链向后顺移一位。这两步之后，让nex重新指向cur后面的节点。（如图）

这个方法同样可以用到反置链表的算法中，很好

-1 -> 1 -> 2 -> 3 -> 4 -> 5
 |    |    |    
pre  cur  nex  

-1 -> 2 -> 1 -> 3 -> 4 -> 5
 |         |    |    
pre       cur  nex  

-1 -> 3 -> 2 -> 1 -> 4 -> 5
 |              |    |    
pre            cur  nex  

https://leetcode.com/problems/reverse-nodes-in-k-group/discuss/

class Solution {
public:
    ListNode *reverseKGroup(ListNode *head, int k) {
        if(head==NULL||k==1) return head;
        int num=0;
        ListNode *preheader = new ListNode(-1);
        preheader->next = head;   //构造带头结点的链表
        ListNode *cur = preheader, *nex, *pre = preheader;
        while(cur = cur->next)   //计算共有多少个节点
            num++;
        while(num>=k) {     
            cur = pre->next;
            nex = cur->next;
            for(int i=1;i<k;++i) {  //把连续k个节点反过来
                cur->next=nex->next;
                nex->next=pre->next;
                pre->next=nex;
                nex=cur->next;
            }
            pre = cur;    //进行下一轮k个节点的操作
            num-=k; 
        }
        return preheader->next; 
    } 
};

看讨论区还有递归方法，现在看不进去了，先会一个方法。感兴趣可以看一下