9-degree OJ1511: print the linked list from the end to the end, offeroj1511

9-degree OJ1511: print the linked list from the end to the end, offeroj1511

Question link:
Http://ac.jobdu.com/problem.php? Pid = 1, 1511

Print the linked list from the end to the end

Time Limit: 1 second memory limit: 128 MB Special Judgment: No submitted: 6036 resolution: 1817
Description:
Enter a chain table to print the values of each node in the chain table from the end to the end.
Input:
Each input file contains only one set of test samples.
Each group of test cases contains multiple rows. Each row contains an integer greater than 0, representing a linked list node. The first line is the value of the first node of the linked list, and so on. When the input time is-1, the Table chain table is input. -1 itself does not belong to the linked list.
Output:
For each test case, the values of each node in the chain table are output from the end to the end, and each value occupies one row.
Sample input:
1
2
3
4
5
-1
Sample output:
5
4
3
2
1

Analysis:

Method 1: generate a linked list in reverse order using the stack
The linked list is always accessed from the beginning during access, while accessing the stack. The stack features advanced post-release, so the reverse input of the linked list is used when the stack is output.
Method 2: Recursive Implementation, perfect
Each time a node is accessed, a node following the output is recursively accessed, and then the node itself is output in reverse order.

Method 2 9-degree AC implementation:

# Include
 
  
# Include
  
   
# Include
   
    
# Include
    
     
Using namespace std; typedef struct Node {int data; struct Node * next;} Node, * pNode;/* print a single-chain table recursively from the end to the end */void PrintListReverse (pNode pHead) {if (pHead = NULL) return; if (pHead-> next! = NULL) PrintListReverse (pHead-> next); printf ("% d \ n", pHead-> data);} pNode CreateList () {int val; pNode pHead = NULL; pNode pCur = NULL; do {scanf ("% d", & val); if (val! =-1) {pNode pNew = (pNode) malloc (sizeof (Node); pNew-> data = val; pNew-> next = NULL; if (pHead = NULL) {pHead = pNew; pCur = pHead;} else {pCur-> next = pNew; pCur = pCur-> next ;}} while (val! =-1); return pHead;} int main () {pNode pHead = CreateList (); PrintListReverse (pHead); return 0 ;} "data-snippet-id =" ext.74a6fec1324c46e2e1780fa1f59e24c8 "data-snippet-saved =" false "data-csrftoken =" sdo5J4AD-xNLccKJgD_R6uZEKjMnYTCN1cP8 "data-codota-status =" done ">
     /******************************** ------------------------------- [Sword refers to the offline interview questions] print the linked list ----------------------------------------- Author: mu zhidian Date: 2015 Email: bzhou84@163.com **********************************/# include <stdio. h> # include <stdlib. h >#include <cstring> # include <string> # include <iostream> using namespace std; typedef struct Node {int data; struct Node * next;} Node, * pNode; /* recursively print a single-chain table from the end to the header */void PrintListReverse (pNode pHead) {if (pHead = NULL) return; if (pHead-> next! = NULL) PrintListReverse (pHead-> next); printf ("% d \ n", pHead-> data);} pNode CreateList () {int val; pNode pHead = NULL; pNode pCur = NULL; do {scanf ("% d", & val); if (val! =-1) {pNode pNew = (pNode) malloc (sizeof (Node); pNew-> data = val; pNew-> next = NULL; if (pHead = NULL) {pHead = pNew; pCur = pHead;} else {pCur-> next = pNew; pCur = pCur-> next ;}} while (val! =-1); return pHead;} int main () {pNode pHead = CreateList (); PrintListReverse (pHead); return 0 ;}
    
   
  
 

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