1 #define IO std::ios::sync_with_stdio(0),cin.tie(0),cout.tie(0) 2 #define bug(x) cout<<#x<<" is "<<x<<endl; 3 #include<bits/stdc++.h> 4 using namespace std; 5 typedef long long ll; 6 struct Node{ 7 int x; 8 Node*next; 9 Node(int x1,Node *next1){ 10 x=x1,next=next1; 11 } 12 }; 13 void reverse(Node *&head){ 14 if(!head||!head->next)return; 15 Node *p=head; 16 Node *cur=nullptr; 17 while(p){ 18 Node *q=p->next; 19 p->next=cur; 20 cur=p; 21 p=q; 22 } 23 head=cur; 24 } 25 Node* reverse1(Node *&pre,Node *&p){ 26 if(!p->next){ 27 p->next=pre; 28 return p; 29 } 30 Node *q=p->next; 31 p->next=pre; 32 pre=p; 33 return reverse1(pre,q); 34 } 35 void print(Node *head){ 36 while(head){ 37 cout<<head->x<<" "; 38 head=head->next; 39 } 40 cout<<endl; 41 } 42 int main(){ 43 Node *head=new Node(1,nullptr); 44 Node *pre=head; 45 for(int i=2;i<=5;i++){ 46 Node *p=new Node(i,nullptr); 47 pre->next=p; 48 pre=p; 49 } 50 print(head); 51 reverse(head); 52 print(head); 53 Node *p=nullptr; 54 head=reverse1(p,head); 55 print(head); 56 p=nullptr; 57 head=reverse1(p,head); 58 print(head); 59 }
标签:Node,pre,head,迭代,递归,next,链表,print,reverse1 From: https://www.cnblogs.com/ccsu-kid/p/18228894