标签：ber example lis public head hang 指针 this nod
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：看起来比较简答的一题，但要注意一些细节，因为有可能要多次旋转pair(pair大小为k)， 注意每个pair之间的链接，pair内部的反转使用三指针反转链表实现
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode* reverseKGroup(ListNode* head, int k) { if(head ==NULL) return head; ListNode **pp = &head; ListNode* begin = head; ListNode* end = begin; while(begin){ end = begin; for(int i=0; end&& i<k-1; i++) end = end->next; if(end ==NULL) break; ListNode* p1=begin; ListNode* p2= p1->next; ListNode* tailNext = end->next; while(p2&& p2!=tailNext){ ListNode* t = p2->next; p2->next = p1; p1= p2; p2 = t; } begin->next = tailNext; *pp = p1; pp = &(begin->next); begin = tailNext; } return head; } };
标签：ber example lis public head hang 指针 this nod
原文地址：http://www.cnblogs.com/willwu/p/6135926.html
