Remove Duplicates from Sorted List

• LeetCode: Remove Duplicates from Sorted List

• Problem:

Given a sorted linked list, delete all duplicates such that each element appear only once.

• Examples:

For example,
Given 1->1->2, return 1->2.
Given 1->1->2->3->3, return 1->2->3.

• Analysis:

Declare two pointers, A and B. 

One of them, A, is to point at a dinstinct number starting node, the other, B,  is to scan the linked list.

While the val of A isn't equal to the val of B, A.next = B, B scan the next node.

• Code - Java:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if (head == null) {
            return null;
        }
        ListNode prev = head;
        ListNode ptr = head.next;
        while (ptr != null) {
            if (ptr.val != prev.val) {
                prev.next = ptr;
                prev = prev.next;
            }
            ptr = ptr.next;
        }
        prev.next = null;
        return head;
    }
}

• Code - C++:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        if (head == NULL) {
            return NULL;
        }
        ListNode* prev = head;
        ListNode* ptr = head->next;
        while (ptr != NULL) {
            if (ptr->val != prev->val) {
                prev->next = ptr;
                prev = prev->next;
            }
            ptr = ptr->next;
        }
        prev->next = NULL;
        return head;
    }
};

• Code - Python:

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def deleteDuplicates(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if head == None:
            return None
        prev, ptr = head, head.next
        while ptr != None:
            if ptr.val != prev.val:
                prev.next = ptr
                prev = prev.next
            ptr = ptr.next
        prev.next = None
        return head

• Code - C:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* deleteDuplicates(struct ListNode* head) {
    if (head == 0) {
        return 0;
    }
    struct ListNode* prev = head;
    struct ListNode* ptr = head->next;
    while (ptr != 0) {
        if (ptr->val != prev->val) {
            prev->next = ptr;
            prev = prev->next;
        }
        ptr = ptr->next;
    }
    prev->next = 0;
    return head;
}

• Time Complexity: O(N), N is the length of listedlist

• Space Complexity: O(1)