优雅地反转链表

反转单向链表的程序.

我给的代码是递归的实现, 运用递归就可以避免很多特殊情况的考虑, 看起来也比较优雅.

 

#include <iostream>#include <deque>#include <assert.h>using namespace std;template <typename T>struct Node{ T data; Node* next;};template <typename T>void createList(Node<T>*& t){ t = new Node<T>(); t->next = NULL; Node<T>* p = t; Node<T>* q = NULL; for(int i = 0; i < 10; ++i) { q = new Node<T>(); p->next = q; q->next = NULL; q->data = i; p = p->next; }}template <typename T>Node<T>* reverseList1(Node<T> *&t, Node<T>* p){ assert(t != NULL); if(p == NULL) return p; if(p->next) { Node<T>* pTemp = reverseList1(t, p->next); pTemp->next = p; p->next = NULL; } else t->next = p; return p;}template <typename T>void reverseList2(Node<T>*& t, Node<T>* p){ assert(t != NULL); if(p == NULL) return; Node<T>* pre = NULL; Node<T>* next = NULL; while(p) { next = p->next; p->next = pre; pre = p; p = next; } t->next = pre;}template <typename T>void printList(Node<T>* t){ Node<T>* p = t->next; while(p) { cout << p->data << ' '; p = p->next; } cout << endl;}int main(){ Node<int>* t = NULL; createList(t); printList(t); reverseList2(t, t->next); printList(t); return 0;} 

Test in CodeBlocks.