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Leetcode[81, 82] Remove Duplicates from Sorted List

06 Nov 2015

###Task1 Given a sorted linked list, delete all duplicates such that each element appear only once.

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

###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 or head.next == None:
            return head
        cur = head
        while head.next:
            if head.val == head.next.val:
                head.next = head.next.next
            else:
                head = head.next
        return cur

###Java


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

###Task2 Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.

For example, Given 1->2->3->3->4->4->5, return 1->2->5. Given 1->1->1->2->3, return 2->3.

###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 or head.next == None:
            return head
        dummy = ListNode(0)
        dummy.next = head
        head = dummy
        while head.next and head.next.next:
            if head.next.val == head.next.next.val:
                cur = head.next.val
                while head.next and head.next.val == cur:
                    head.next = head.next.next
            else:
                head = head.next
            
        return dummy.next

###Java


/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        //edge
        if (head == null) {
            return null;
        }
        
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        head = dummy;
        while (head.next != null && head.next.next != null) {
            if (head.next.val == head.next.next.val) {
                int value = head.next.val;
                while (head.next != null && head.next.val == value) {
                    head.next = head.next.next;
                }
            } else {
                head = head.next;
            }
        }
        
        return dummy.next;
    }
}

###Points

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