首页 > 代码库 > Reverse Nodes in k-Group
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.
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 };
Reverse Nodes in k-Group
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。