Leetcode and Lintcode

Reverse Linked List II

Problem:s

Reverse a linked list from position m to n.

*Example*
Given 1->2->3->4->5->NULL, m = 2 and n = 4, return 1->4->3->2->5->NULL.

Note
Given m, n satisfy the following condition: 1 ≤ m ≤ n ≤ length of list.

Challenge
Reverse it in-place and in one-pass

Leetcode link
Lintcode link

Solution:

This problem is complex which has many edge cases so that it is not easy to write a bug free program within a short time.
The idea of the solution is to go through the list, to the m, store the previous node before m, and start reversing from m to n.
Then link the non-reversed and reversed part together.

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/*************************************************************************
	> File Name: ReverseLinkedListII.java
	> Author: Yao Zhang 
	> Mail: psyyz10@163.com 
	> Created Time: Mon 16 Nov 14:48:56 2015
 ************************************************************************/

public class ReverseLinkedListII{
    // Solution 1
    public static ListNode reverseBetween(ListNode head, int m, int n) {
        if (m>=n) return head;
        
        ListNode current = head;
        int i;
        ListNode startNode = null, newNode = null, previousNode = null;
        for (i = 1; i <= n; i++){
            if (i == m - 1){
                previousNode = current;
            } 
            if (i == m){
                 startNode = newNode =  current;
            }
            
            if (i>m){
                ListNode temp = current.next;
                current.next = newNode;
                newNode = current;
                current = temp;
                continue;
            }
            
            current = current.next;
        }
        
        startNode.next = current; 
        
        if (previousNode == null) head = newNode;
        else previousNode.next = newNode;
        
        return head;
    }

    // Solution 2
    public ListNode reverseBetween(ListNode head, int m, int n) {
        if (m >= n || head == null)
            return head;
            
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        ListNode prev = dummy;
                                
        for (int i = 0; i < m - 1 ; i++){
            if (prev == null) {
                return dummy.next;
            }
            prev = prev.next;  
        }                       
        
        ListNode mPrev = prev;
        ListNode current = prev.next;
        ListNode last = new ListNode(current.val);
        ListNode first = last;            
        current = current.next;
                                
        for (int i = 0; i < n - m ; i++){
            if (current == null) {
                return dummy.next;
            }
            ListNode temp = current.next;
            current.next = first;
            first = current;
            current = temp;
        }                       
                                
        mPrev.next = first;
                                
        last.next = current; 
                                
        return dummy.next;            
    }
}