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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:其实就是反转多个链表,然后把这多个链表连接起来即可.
code:
class Solution { public: ListNode *reverseKGroup(ListNode *head, int k) { if(head==NULL||k<0) return NULL; if(k==0||head->next==NULL) return head; bool first=true; vector<ListNode*> lastOfGroup;int loc=0; ListNode* prep=NULL; ListNode* p=NULL; ListNode* move=head; ListNode* nextHead=NULL; ListNode* resHead=NULL; bool firsttag=true; while(move!=NULL){ int i=1; nextHead=move; while(i<=k&&move!=NULL){ move=move->next; ++i; } prep=nextHead; p=nextHead->next; lastOfGroup.push_back(prep); if(i-1==k){ prep->next=NULL; int j=1; while(j<=k-1){ ListNode* tempNextNode=p->next; p->next=prep; prep=p; p=tempNextNode; ++j; } if(firsttag) resHead=prep; if(!firsttag) lastOfGroup[loc++]->next=prep; firsttag=false; }else{ if(firsttag) return head; lastOfGroup[loc++]->next=nextHead; } } return resHead; } };
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原文地址：http://www.cnblogs.com/fightformylife/p/4218778.html
