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How to sort ListNode (linked) by its value?
How do I check if an array includes an object in JavaScript?How to detect a loop in a linked list?Fastest sort of fixed length 6 int arraycheck if all elements in a list are identicalInterview Question: Merge two sorted singly linked lists without creating new nodesHow to pair socks from a pile efficiently?Is there a way to measure how sorted a list is?removing even indexes' values in a linked listWhat's wrong with my code to delete duplicate-value nodes from a sorted linked list?Linked list merge sort using java
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How do you sort nodelist by value without using any library.
Example:
* Input: 3->1->5->4->2
* Output: 1->2->3->4->5
ListNode.java
import java.util.List;
public class ListNode
public int val;
public ListNode next;
public ListNode(int x)
val = x;
public ListNode(int val, ListNode next)
this.val = val;
this.next = next;
SortLinkList.java
public class SortLinkList
public static ListNode sortLinkList(ListNode list)
//TODO:
return list;
```
algorithm linked-list
edited Mar 23 at 10:43
Mark Setchell
94.7k786197
asked Mar 23 at 7:31
The Java GuyThe Java Guy
92478
add a comment |
How do you sort nodelist by value without using any library.
Example:
* Input: 3->1->5->4->2
* Output: 1->2->3->4->5
ListNode.java
import java.util.List;
public class ListNode
public int val;
public ListNode next;
public ListNode(int x)
val = x;
public ListNode(int val, ListNode next)
this.val = val;
this.next = next;
SortLinkList.java
public class SortLinkList
public static ListNode sortLinkList(ListNode list)
//TODO:
return list;
```
algorithm linked-list
edited Mar 23 at 10:43
Mark Setchell
94.7k786197
asked Mar 23 at 7:31
The Java GuyThe Java Guy
92478
add a comment |
How do you sort nodelist by value without using any library.
Example:
* Input: 3->1->5->4->2
* Output: 1->2->3->4->5
ListNode.java
import java.util.List;
public class ListNode
public int val;
public ListNode next;
public ListNode(int x)
val = x;
public ListNode(int val, ListNode next)
this.val = val;
this.next = next;
SortLinkList.java
public class SortLinkList
public static ListNode sortLinkList(ListNode list)
//TODO:
return list;
```
algorithm linked-list
edited Mar 23 at 10:43
Mark Setchell
94.7k786197
asked Mar 23 at 7:31
The Java GuyThe Java Guy
92478
How do you sort nodelist by value without using any library.
Example:
* Input: 3->1->5->4->2
* Output: 1->2->3->4->5
ListNode.java
import java.util.List;
public class ListNode
public int val;
public ListNode next;
public ListNode(int x)
val = x;
public ListNode(int val, ListNode next)
this.val = val;
this.next = next;
SortLinkList.java
public class SortLinkList
public static ListNode sortLinkList(ListNode list)
//TODO:
return list;
```
algorithm linked-list
algorithm linked-list
edited Mar 23 at 10:43
Mark Setchell
94.7k786197
asked Mar 23 at 7:31
The Java GuyThe Java Guy
92478
edited Mar 23 at 10:43
Mark Setchell
94.7k786197
asked Mar 23 at 7:31
The Java GuyThe Java Guy
92478
edited Mar 23 at 10:43
Mark Setchell
94.7k786197
edited Mar 23 at 10:43
Mark Setchell
94.7k786197
edited Mar 23 at 10:43
Mark Setchell
94.7k786197
94.7k786197
asked Mar 23 at 7:31
The Java GuyThe Java Guy
92478
asked Mar 23 at 7:31
The Java GuyThe Java Guy
92478
asked Mar 23 at 7:31
The Java GuyThe Java Guy
92478
92478
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
If you want in-place sort you can just implement bubble-sort:
Pseudo-code:
bool notDone = true
while(notDone)
notDone = false;
cur = head;
while(cur.nxt != null)
prev = cur;
cur = cur.nxt;
if(cur.val > cur.nxt.val)
prev.nxt = cur.nxt;
temp = cur.nxt.nxt;
cur.nxt.nxt = cur;
cur.nxt = temp;
notDone = true;
answered Mar 23 at 7:49
PhotonPhoton
827814
The concept is right however it doesn't sort the list.
– The Java Guy
Mar 24 at 9:44
add a comment |
1.Make an array of the class which only store each node and for each node, next is pointed to null.Length of the array is no of nodes in the list.
2.Sort the array
3. Link the nodes and return head
PS: why are you using linked list when there is a sorting operation involved , instead of using arrays alone?
answered Mar 23 at 7:50
Argha ChakrabortyArgha Chakraborty
259
This is algo problem. want to solve without libraries.
– The Java Guy
Mar 24 at 12:49
The solution that is given here does not require any library. I'm assuming you know how to sort (using well known sorting algorithm)an array of objects based on an attribute? Each object being the node of your list.
– Argha Chakraborty
Mar 24 at 13:25
Your assumption is right. This will require extra O(n) space + O(n^2) worst case runtime depending on which sorting you use. This is not something that I am using in my production code. This aas one of the algorithm question.
– The Java Guy
Mar 25 at 20:40
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
If you want in-place sort you can just implement bubble-sort:
Pseudo-code:
bool notDone = true
while(notDone)
notDone = false;
cur = head;
while(cur.nxt != null)
prev = cur;
cur = cur.nxt;
if(cur.val > cur.nxt.val)
prev.nxt = cur.nxt;
temp = cur.nxt.nxt;
cur.nxt.nxt = cur;
cur.nxt = temp;
notDone = true;
answered Mar 23 at 7:49
PhotonPhoton
827814
The concept is right however it doesn't sort the list.
– The Java Guy
Mar 24 at 9:44
add a comment |
If you want in-place sort you can just implement bubble-sort:
Pseudo-code:
bool notDone = true
while(notDone)
notDone = false;
cur = head;
while(cur.nxt != null)
prev = cur;
cur = cur.nxt;
if(cur.val > cur.nxt.val)
prev.nxt = cur.nxt;
temp = cur.nxt.nxt;
cur.nxt.nxt = cur;
cur.nxt = temp;
notDone = true;
answered Mar 23 at 7:49
PhotonPhoton
827814
The concept is right however it doesn't sort the list.
– The Java Guy
Mar 24 at 9:44
add a comment |
If you want in-place sort you can just implement bubble-sort:
Pseudo-code:
bool notDone = true
while(notDone)
notDone = false;
cur = head;
while(cur.nxt != null)
prev = cur;
cur = cur.nxt;
if(cur.val > cur.nxt.val)
prev.nxt = cur.nxt;
temp = cur.nxt.nxt;
cur.nxt.nxt = cur;
cur.nxt = temp;
notDone = true;
answered Mar 23 at 7:49
PhotonPhoton
827814
If you want in-place sort you can just implement bubble-sort:
Pseudo-code:
bool notDone = true
while(notDone)
notDone = false;
cur = head;
while(cur.nxt != null)
prev = cur;
cur = cur.nxt;
if(cur.val > cur.nxt.val)
prev.nxt = cur.nxt;
temp = cur.nxt.nxt;
cur.nxt.nxt = cur;
cur.nxt = temp;
notDone = true;
answered Mar 23 at 7:49
PhotonPhoton
827814
answered Mar 23 at 7:49
PhotonPhoton
827814
answered Mar 23 at 7:49
PhotonPhoton
827814
answered Mar 23 at 7:49
PhotonPhoton
827814
827814
The concept is right however it doesn't sort the list.
– The Java Guy
Mar 24 at 9:44
add a comment |
The concept is right however it doesn't sort the list.
– The Java Guy
Mar 24 at 9:44
The concept is right however it doesn't sort the list.
– The Java Guy
Mar 24 at 9:44
The concept is right however it doesn't sort the list.
– The Java Guy
Mar 24 at 9:44
add a comment |
1.Make an array of the class which only store each node and for each node, next is pointed to null.Length of the array is no of nodes in the list.
2.Sort the array
3. Link the nodes and return head
PS: why are you using linked list when there is a sorting operation involved , instead of using arrays alone?
answered Mar 23 at 7:50
Argha ChakrabortyArgha Chakraborty
259
This is algo problem. want to solve without libraries.
– The Java Guy
Mar 24 at 12:49
The solution that is given here does not require any library. I'm assuming you know how to sort (using well known sorting algorithm)an array of objects based on an attribute? Each object being the node of your list.
– Argha Chakraborty
Mar 24 at 13:25
Your assumption is right. This will require extra O(n) space + O(n^2) worst case runtime depending on which sorting you use. This is not something that I am using in my production code. This aas one of the algorithm question.
– The Java Guy
Mar 25 at 20:40
add a comment |
1.Make an array of the class which only store each node and for each node, next is pointed to null.Length of the array is no of nodes in the list.
2.Sort the array
3. Link the nodes and return head
PS: why are you using linked list when there is a sorting operation involved , instead of using arrays alone?
answered Mar 23 at 7:50
Argha ChakrabortyArgha Chakraborty
259
This is algo problem. want to solve without libraries.
– The Java Guy
Mar 24 at 12:49
The solution that is given here does not require any library. I'm assuming you know how to sort (using well known sorting algorithm)an array of objects based on an attribute? Each object being the node of your list.
– Argha Chakraborty
Mar 24 at 13:25
Your assumption is right. This will require extra O(n) space + O(n^2) worst case runtime depending on which sorting you use. This is not something that I am using in my production code. This aas one of the algorithm question.
– The Java Guy
Mar 25 at 20:40
add a comment |
1.Make an array of the class which only store each node and for each node, next is pointed to null.Length of the array is no of nodes in the list.
2.Sort the array
3. Link the nodes and return head
PS: why are you using linked list when there is a sorting operation involved , instead of using arrays alone?
answered Mar 23 at 7:50
Argha ChakrabortyArgha Chakraborty
259
1.Make an array of the class which only store each node and for each node, next is pointed to null.Length of the array is no of nodes in the list.
2.Sort the array
3. Link the nodes and return head
PS: why are you using linked list when there is a sorting operation involved , instead of using arrays alone?
answered Mar 23 at 7:50
Argha ChakrabortyArgha Chakraborty
259
answered Mar 23 at 7:50
Argha ChakrabortyArgha Chakraborty
259
answered Mar 23 at 7:50
Argha ChakrabortyArgha Chakraborty
259
answered Mar 23 at 7:50
Argha ChakrabortyArgha Chakraborty
259
259
This is algo problem. want to solve without libraries.
– The Java Guy
Mar 24 at 12:49
The solution that is given here does not require any library. I'm assuming you know how to sort (using well known sorting algorithm)an array of objects based on an attribute? Each object being the node of your list.
– Argha Chakraborty
Mar 24 at 13:25
Your assumption is right. This will require extra O(n) space + O(n^2) worst case runtime depending on which sorting you use. This is not something that I am using in my production code. This aas one of the algorithm question.
– The Java Guy
Mar 25 at 20:40
add a comment |
This is algo problem. want to solve without libraries.
– The Java Guy
Mar 24 at 12:49
The solution that is given here does not require any library. I'm assuming you know how to sort (using well known sorting algorithm)an array of objects based on an attribute? Each object being the node of your list.
– Argha Chakraborty
Mar 24 at 13:25
Your assumption is right. This will require extra O(n) space + O(n^2) worst case runtime depending on which sorting you use. This is not something that I am using in my production code. This aas one of the algorithm question.
– The Java Guy
Mar 25 at 20:40
This is algo problem. want to solve without libraries.
– The Java Guy
Mar 24 at 12:49
This is algo problem. want to solve without libraries.
– The Java Guy
Mar 24 at 12:49
The solution that is given here does not require any library. I'm assuming you know how to sort (using well known sorting algorithm)an array of objects based on an attribute? Each object being the node of your list.
– Argha Chakraborty
Mar 24 at 13:25
The solution that is given here does not require any library. I'm assuming you know how to sort (using well known sorting algorithm)an array of objects based on an attribute? Each object being the node of your list.
– Argha Chakraborty
Mar 24 at 13:25
Your assumption is right. This will require extra O(n) space + O(n^2) worst case runtime depending on which sorting you use. This is not something that I am using in my production code. This aas one of the algorithm question.
– The Java Guy
Mar 25 at 20:40
Your assumption is right. This will require extra O(n) space + O(n^2) worst case runtime depending on which sorting you use. This is not something that I am using in my production code. This aas one of the algorithm question.
– The Java Guy
Mar 25 at 20:40
add a comment |
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