203_Removed-Linked-List-Elements

203_Removed-Linked-List-Elements

Description

Remove all elements from a linked list of integers that have value val.

Example:

Input:  1->2->6->3->4->5->6, val = 6
Output: 1->2->3->4->5

Solution

Java solution 1

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode removeElements(ListNode head, int val) {
        while (head != null && head.val == val) {
            head = head.next;
        }

        if (head == null) {
            return null;
        }

        ListNode prev = head;
        while (prev.next != null) {
            if (prev.next.val == val) {
                prev.next = prev.next.next;
            } else {
                prev = prev.next;
            }
        }

        return head;
    }
}

Runtime: 7 ms.

Java solution 2

Using dummy head node.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode removeElements(ListNode head, int val) {
        ListNode dummyHead = new ListNode(-1);
        dummyHead.next = head;

        ListNode prev = dummyHead;
        while (prev.next != null) {
            if (prev.next.val == val) {
                prev.next = prev.next.next;
            } else {
                prev = prev.next;
            }
        }

        return dummyHead.next;
    }
}

Runtime: 8 ms.

Python solution

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

class Solution:
    def removeElements(self, head, val):
        """
        :type head: ListNode
        :type val: int
        :rtype: ListNode
        """
        dummy_head = ListNode(-1)
        dummy_head.next = head
        
        prev = dummy_head
        while prev.next is not None:
            if prev.next.val == val:
                prev.next = prev.next.next
            else:
                prev = prev.next
                
        return dummy_head.next

Runtime: 88 ms. Your runtime beats 74.70 % of python3 submissions.

posted @ 2018-06-25 10:09  xugenpeng  阅读(167)  评论(0编辑  收藏  举报