Description:
Given the head of a sorted linked list, delete all duplicates such that each element appears only once. Return the linked list sorted as well.
Examples:
Example 1:
Input: head = [1,1,2]
Output: [1,2]
Example 2:
Input: head = [1,1,2,3,3]
Output: [1,2,3]Solution in Python:
Python
# Definition for singly-linked list.
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
class Solution:
def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:
# Initialize current node as the head of the list
current = head
# Traverse the linked list until the end
while current and current.next:
# If the current node's value is the same as the next node's value
if current.val == current.next.val:
# Skip the next node by pointing the current node's next to the next node's next
current.next = current.next.next
else:
# Move to the next node
current = current.next
# Return the modified list's head
return headExplanation:
- ListNode Class:
- This is the definition of a singly-linked list node with a value (
val) and a pointer to the next node (next).
- This is the definition of a singly-linked list node with a value (
- Current Pointer:
current: A pointer to traverse the linked list starting from the head.
- Traversal and Duplicate Removal:
- We traverse the list with the
currentpointer until we reach the end of the list (while current and current.next). - For each node, we check if its value is the same as the value of the next node (
current.val == current.next.val). - If they are the same, we skip the next node by setting the
current.nextpointer tocurrent.next.next, effectively removing the duplicate node from the list. - If they are not the same, we simply move the
currentpointer to the next node.
- We traverse the list with the
- Return Result:
- Finally, we return the head of the modified list, which now contains only unique elements in sorted order.
Solution in Javascript:
JavaScript
/**
* Definition for singly-linked list.
* function ListNode(val, next) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {ListNode} head
* @return {ListNode}
*/
var deleteDuplicates = function(head) {
// Initialize current node as the head of the list
let current = head;
// Traverse the linked list until the end
while (current !== null && current.next !== null) {
// If the current node's value is the same as the next node's value
if (current.val === current.next.val) {
// Skip the next node by pointing the current node's next to the next node's next
current.next = current.next.next;
} else {
// Move to the next node
current = current.next;
}
}
// Return the modified list's head
return head;
};Solution in Java:
Java
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode deleteDuplicates(ListNode head) {
// Initialize the current node as the head of the list
ListNode current = head;
// Traverse the linked list until the end
while (current != null && current.next != null) {
// If the current node's value is the same as the next node's value
if (current.val == current.next.val) {
// Skip the next node by pointing the current node's next to the next node's next
current.next = current.next.next;
} else {
// Move to the next node
current = current.next;
}
}
// Return the modified list's head
return head;
}
}Solution in C#:
C#
/**
* Definition for singly-linked list.
* public class ListNode {
* public int val;
* public ListNode next;
* public ListNode(int val=0, ListNode next=null) {
* this.val = val;
* this.next = next;
* }
* }
*/
public class Solution {
public ListNode DeleteDuplicates(ListNode head) {
// Initialize the current node as the head of the list
ListNode current = head;
// Traverse the linked list until the end
while (current != null && current.next != null) {
// If the current node's value is the same as the next node's value
if (current.val == current.next.val) {
// Skip the next node by pointing the current node's next to the next node's next
current.next = current.next.next;
} else {
// Move to the next node
current = current.next;
}
}
// Return the modified list's head
return head;
}
}