出个题目,O(n)时间里逆转单链表 补全函数。
补写reverse函数,实现链表逆转。
递归的,非递归的。
程序代码:
#include<iostream> using namespace std; struct Exception{ Exception(){cout<<"Exception"<<endl;} }; template<class Type> struct node{ Type val; node<Type> *next; node<Type>(){next=NULL;} node(Type v):val(v){next=NULL; } friend ostream & operator<<(ostream &out,node<Type> & n) { out<<n.val; return out; } }; template <class Type> class SList{ private: node<Type> *hummy; protected: node<Type> * getPre(node<Type> *pointer) { try{ node<Type> * n; for(n=hummy;n!=NULL;n=n->next) { if( n->next==pointer) break; } if( pointer== hummy) throw Exception(); return n; }catch( Exception () ) { cout<<"Fist node with out pre node"<<endl; return NULL; } } node<Type> * getNext(node<Type> *pointer) { try{ if(pointer->next==NULL) throw Exception (); return pointer->next; }catch( Exception e) { cout<<"End of List"<<endl; return NULL; } } bool insert(node<Type> *pointer, Type v) { try { node<Type> *n=new node<Type>(v); if( NULL == n ) throw Exception(); else { n->next=pointer->next; pointer->next=n; return true; } }catch (Exception e ) { cout<<"New Error"<<endl; return false; } } bool del(node<Type> *pointer) { try { if( pointer != hummy->next) { node<Type> * pre=getPre(pointer); pre->next=pointer->next; delete pointer; pointer=NULL; } else { hummy->next=pointer->next; delete pointer; pointer=NULL; } }catch( Exception()) { } } public: SList(){ hummy=new node<Type>();} SList(int n) { hummy=new node<Type>(); for( int i=0;i<n;i++) { Type m; cin>>m; insert(hummy,m); } } friend void reverse( SList & s) { } void travel() { node<Type> * n; for( n=hummy->next; n!=NULL;n=n->next) { cout<<*n<<" "; } cout<<endl; } }; int main() { int m; cin>>m; SList<int> sl(m); sl.travel(); reverse(sl); sl.travel(); return 0; }