Consider the following code for reversing a singly linked list. Fill in the blank with Java code to complete the method implementation. Write your answers with no spaces. public void reverse() { } Node p = head, prev = null; tail = p; while (answer1_) { Node next = p.next; answer2___; prev = p; } head = prev; answer1 = answer2 = answer3___; answer3 =
Consider the following code for reversing a singly linked list. Fill in the blank with Java code to complete the method implementation. Write your answers with no spaces. public void reverse() { } Node p = head, prev = null; tail = p; while (answer1_) { Node next = p.next; answer2___; prev = p; } head = prev; answer1 = answer2 = answer3___; answer3 =
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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Question

Transcribed Image Text:### Reversing a Singly Linked List in Java
**Task Description:**
Consider the following Java code for reversing a singly linked list. Fill in the blanks with Java code to complete the method implementation. Write your answers with no spaces.
```java
public void reverse() {
Node p = head, prev = null;
tail = p;
while (__answer1__) {
Node next = p.next;
__answer2__;
prev = p;
__answer3__;
}
head = prev;
}
```
**Answer Fields:**
- **answer1 =** (condition to continue looping)
- **answer2 =** (statement to reverse the current node pointer)
- **answer3 =** (statement to move to the next node)
To solve this, identify the logic needed to reverse the pointers in the linked list, ensuring to specify the loop condition, pointer reversal, and the traversal to the next node.
Expert Solution

Step 1
The code is given for reversing the singly linked list. For reversing the list, there's is three pointers used (p, prev, tail). Then iterate through the linked list.
The code given is:-
public void reverse() {
Node p = head, prev = null;
tail = p;
while (answer1) {
Node next = p.next;
answer2_;
prev = p;
answer3_;
}
head = prev;
}
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