首页 > 代码库 > Leetcode: Reverse Nodes in k-Group
Leetcode: 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
分析：每k个node为一组做reverse，时间复杂度O(n), 空间复杂度O(1).
class Solution {public: ListNode *reverseKGroup(ListNode *head, int k) { if(head == NULL || head->next == NULL || k == 0 || k == 1) return head; ListNode *dummy = new ListNode(-1); dummy->next = head; ListNode *pre_f = dummy, *first = head, *second = head, *post_s = head->next; while(first){ int c = 0; while(second && c < k-1){ second = second->next; post_s = second?second->next:NULL; c++; } if(second){//c == k-1 second->next = NULL; pre_f->next = reverse(first); first->next = post_s; pre_f = first; first = post_s; second = post_s; }else break; } return dummy->next; } ListNode * reverse(ListNode * head){ if(head == NULL || head->next == NULL) return head; ListNode *dummy = new ListNode(-1); dummy->next = head; ListNode *p = head; while(p->next){ ListNode *tmp = p->next; p->next = p->next->next; tmp->next = dummy->next; dummy->next = tmp; } return dummy->next; }};
Leetcode: 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。