LeetCode – Reverse Linked List II (Java)

Category: Algorithms >> Interview June 5, 2014

Reverse a linked list from position m to n. Do it in-place and in one-pass.

For example: given 1->2->3->4->5->NULL, m = 2 and n = 4, return 1->4->3->2->5->NULL.

Analysis

Java Solution

public ListNode reverseBetween(ListNode head, int m, int n) {
    if(m==n) return head;
 
    ListNode prev = null;//track (m-1)th node
    ListNode first = new ListNode(0);//first's next points to mth
    ListNode second = new ListNode(0);//second's next points to (n+1)th
 
    int i=0;
    ListNode p = head;
    while(p!=null){
        i++;
        if(i==m-1){
            prev = p;
        }
 
        if(i==m){
            first.next = p;
        }
 
        if(i==n){
            second.next = p.next;
            p.next = null;
        }
 
        p= p.next;
    }
    if(first.next == null)
        return head;
 
    // reverse list [m, n]    
    ListNode p1 = first.next;
    ListNode p2 = p1.next;
    p1.next = second.next;
 
    while(p1!=null && p2!=null){
        ListNode t = p2.next;
        p2.next = p1;
        p1 = p2;
        p2 = t;
    }
 
    //connect to previous part
    if(prev!=null)
        prev.next = p1;
    else
        return p1;
 
    return head;
}

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Venk

Simple and best solution found in LeetCode discuss

public static ListNode reverseBetween(ListNode head, int m, int n) {
if(head == null) return null; ListNode dummy = new ListNode(0); // create a dummy node to mark the head of this list dummy.next = head; ListNode pre = dummy; // make a pointer pre as a marker for the node before reversing for(int i = 0; i pre = 1, start = 2, then = 3 // dummy-> 1 -> 2 -> 3 -> 4 -> 5 for(int i=0; i1 - 3 - 2 - 4 - 5; pre = 1, start = 2, then = 4 // second reversing: dummy->1 - 4 - 3 - 2 - 5; pre = 1, start = 2, then = 5 (finish) return dummy.next;
}
Sruthi

public class ReverseLinkedListMtoN2 {

public ListNode reverseBetween(ListNode head, int m, int n) {

if (m != n) {

ListNode curr = head;

int i = 1;

ListNode prevHead = null;

ListNode tail = null;

ListNode prev = null;

ListNode next = null;

while (i < m) {

prevHead = curr;

curr = curr.next;

i++;

}

tail = curr;

next = curr.next;

while (i < n) {

prev = curr;

curr = next;

next = curr.next;

curr.next = prev;

i++;

}

if (m == 1)

head = curr;

else

prevHead.next = curr;

tail.next = next;

}

return head;

}
Shivendra

Why to create new ListNode???
When the same stuff can be done just by rearranging the references?

/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode reverseBetween(ListNode head, int m, int n) {

if(m==n)
return head;

ListNode mprev = null;
ListNode nafter = null;
ListNode mprevprev = null;
ListNode temp = head;

int i=1;
while(temp!=null && i<=n){

if(i==m-1){
mprevprev = temp;
}
if(i==m){
mprev = temp;
}
if(i==n){
nafter = temp.next;
temp.next = null;
}
++i;
temp = temp.next;

}

if(mprev.next == null)
return head;

// Reverse the link list from m to n

ListNode x = mprev;
ListNode y = x.next;
x.next = nafter;

while(x!=null && y!=null){
ListNode t = y.next;
y.next = x;
x=y;
y=t;
}

if(mprevprev!=null)
mprevprev.next =x;
else
return x;

return head;

}
}

LeetCode – Reverse Linked List II (Java)

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