标签:链表的k子段逆转
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; }
标签:链表的k子段逆转
原文地址:http://blog.csdn.net/cjc211322/article/details/39077621