标签：des style blog color io ar for div sp
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
思路：依次处理每k个节点。每次先找到当前处理的K个节点的尾指针，从而获得下一个K个节点的起始地址；然后将当前的k个节点逆序。循环直至当前处理的节点数不足K为止。
1 class Solution { 2 public: 3 ListNode *reverseKGroup( ListNode *head, int k ) { 4 if( k <= 1 ) { return head; } 5 ListNode guard( -1 ); 6 guard.next = head; 7 head = &guard; 8 while( head ) { 9 ListNode *end = head->next; 10 for( int i = 0; i < k; ++i ) { 11 if( !end ) { return guard.next; } 12 end = end->next; 13 } 14 ListNode *node = head->next->next; 15 head->next->next = end; 16 end = head->next; 17 for( int i = 1; i < k; ++i ) { 18 ListNode *tmp = node->next; 19 node->next = head->next; 20 head->next = node; 21 node = tmp; 22 } 23 head = end; 24 } 25 return guard.next; 26 } 27 };
标签：des style blog color io ar for div sp
原文地址：http://www.cnblogs.com/moderate-fish/p/3959211.html
