算法导论10.2-7习题解答(单链表逆转)

CLRS 10.2-7 ：
在O(n)时间内对给定的单链表进行逆转。

解法：

这题没什么技巧，就是将指针的指向进行改变即可。

#include <iostream>
usingnamespace std;

//很简单的Node 和 List
class Node
{
public:
  Node* next;
      int data;

  Node();
  Node(int d);
};

class List
{
public:
  Node* first;

  List();
      void insert(int d);
};

int main()
{
      int K =5;
  List* list  =new List();
    
      for(int i =0; i <5; i++)
  {
    list->insert(i);
  }

  Node* before = list->first;
  Node* after = before->next;
  before->next = NULL;
      while(after)
  {
            //交换指针
    Node* temp = after->next;
    after->next = before;
    before = after;
    after = temp;
  }
      //把最后的before赋给list->first
  list->first = before;

  Node* x = list->first;
      while(x)
  {
    cout<<x->data<<endl;
    x = x->next;
  }
      return0;
}

Node::Node()
{
  next = NULL;
}

Node::Node(int d)
{
  data = d;
  next = NULL;
}

List::List()
{
  first = NULL;
}
//插入节点的操作
void List::insert(int d)
{
  Node* node =new Node(d);

      if(first == NULL)
    first = node;
      else
  {
    Node* temp = first->next;
    Node* prev = first;
            while(temp != NULL)
    {
      prev = temp;
      temp = temp->next;
    }
    prev->next = node;
  }
}