Reverse a linked list

原创转载请注明出处：http://agilestyle.iteye.com/blog/2360694

Recursive Idea

Reverse(Head -> Remaining List)

=>

Reverse(Remaining List) -> Head

Example:

Reverse(1->2->3->4->5)

=>

Reverse(2->3->4->5) -> 1

Demo

package org.fool.java.collections;

public class ReverseLinkedListTest {
    public static void main(String[] args) {
        List l = new List(1);
        l.next = new List(2);
        l.next.next = new List(3);
        l.next.next.next = new List(4);
        l.next.next.next.next = new List(5);

        System.out.println("Original List: " + l.toString());

        l = reverse(l);

        System.out.println("Reversed List: " + l.toString());
    }

    public static List reverse(List l) {
        // firstly check if l is empty or only has one element then return
        if (l == null || l.next == null) {
            return l;
        }

        // otherwise, use recursive method to process
        List remainingReverse = reverse(l.next);

        // step 1: need to update the tail of remaining reverse as head l
        l.next.next = l;    // this (l.next) is the key to get the tail in constant time

        // set l.next to NULL after that! Otherwise it's causing cycles in list
        l.next = null;

        // step 2: return the reverse list
        return remainingReverse;
    }
}

class List {
    int value;
    List next;

    public List(int value) {
        this.value = value;
    }

    @Override
    public String toString() {
        List current = this;
        String output = "";

        while (current != null) {
            output += current.value + " -> ";
            current = current.next; // increment the pointer index current
        }

        return output + "NULL";
    }
}

Note： 

Console Output

 

Reference

https://www.youtube.com/watch?v=j5m6rRszzEQ&list=PLlhDxqlV_-vkak9feCSrnjlrnzzzcopSG