Reverse Nodes in k-Group leetcode java

題目：

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

 

題解：

這道題主要是利用reverse鏈表的方法，reverse的方法就是維護三個指針，然后別忘了保存next指針就行。

代碼如下：

 
           1      
          // 
          http://www.cnblogs.com/lichen782/p/leetcode_Reverse_Nodes_in_kGroup.html 
          
 
           2  
               
          /** 
          
 
           3  
               * Reverse a link list between pre and next exclusively
 
           4  
               * an example:
 
           5  
               * a linked list:
 
           6  
               * 0->1->2->3->4->5->6
 
           7  
               * |           |   
 
           8  
               * pre        next
 
           9  
               * after call pre = reverse(pre, next)
 
          10  
               * 
 
          11  
               * 0->3->2->1->4->5->6
 
          12  
               *          |  |
 
          13  
               *          pre next
 
          14  
               *  
          @param 
           pre 
 
          15  
               *  
          @param 
           next
 
          16  
               *  
          @return 
           the reversed list's last node, which is the precedence of parameter next
 
          17  
                
          */ 
          
 
          18      
          private  
          static ListNode reverse(ListNode pre, ListNode next){ 
          
 
          19         ListNode last = pre.next; 
          // 
          where first will be doomed "last" 
          
 
          20  
                  ListNode cur = last.next; 
          
 
          21          
          while(cur != next){ 
          
 
          22             last.next = cur.next; 
          
 
          23             cur.next = pre.next; 
          
 
          24             pre.next = cur; 
          
 
          25             cur = last.next; 
          
 
          26         } 
          
 
          27          
          return last; 
          
 
          28     } 
          
 
          29      
          
 
          30      
          public  
          static ListNode reverseKGroup(ListNode head,  
          int k) { 
          
 
          31              
          if(head ==  
          null || k == 1) 
          
 
          32                  
          return head; 
          
 
          33                  
          
 
          34             ListNode dummy =  
          new ListNode(0); 
          
 
          35             dummy.next = head; 
          
 
          36              
          int count = 0; 
          
 
          37             ListNode pre = dummy; 
          
 
          38             ListNode cur = head; 
          
 
          39              
          while(cur !=  
          null){ 
          
 
          40                 count ++; 
          
 
          41                 ListNode next = cur.next; 
          
 
          42                  
          if(count == k){ 
          
 
          43                     pre = reverse(pre, next); 
          
 
          44                     count = 0;    
          
 
          45                 } 
          
 
          46                 cur = next; 
          
 
          47             } 
          
 
          48           
          return dummy.next; 
          
 
          49         } 
         

 Reference：http://codeganker.blogspot.com/2014/02/reverse-nodes-in-k-group-leetcode.html