链表的k子段逆转

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题目：

原链表为1->2->3->4->5->6->7

k=3的子段逆转结果为

3->2->1->6->5->4->7

code:

关键点
首子段要记录整个链表的头指针，记录相邻两字段的尾节点

#include "stdafx.h"#include <vector>#include <iostream>#include <hash_map>using namespace std;typedef struct node{int data;struct node* next;}*pList,Node;void rReverse(pList &list,int k){pList p=list, tail, pretail;tail=p;int flag=0;while(p!=NULL){int n=k;pList post=p->next;tail=p;while(post!=NULL && n--!=1){pList temp=post->next;post->next=p;p=post;post=temp;}if(post!=NULL && flag==0){list=p;pretail=tail;flag=1;p=post;}else if(post!=NULL){pretail->next=p;pretail=tail;p=post;}else{pretail->next=p;tail->next=NULL;break;}}}int s[7]={1,2,3,4,5,6,7};void createList(pList& head){head=new Node;pList p=head;for(int i=0;i<7;i++){p->next=new Node;p->next->data=s[i];p=p->next;}p->next=NULL;p=head;head=head->next;delete p;}void showList(pList head){pList p=head;while(p!=NULL){cout<<p->data<<" ";p=p->next;}cout<<endl;}int main(void){pList head;createList(head);showList(head);rReverse(head,4);showList(head);system("pause");return 0;}

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