[LeetCode] Remove Linked List Elements

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

Example
Given: 1 –> 2 –> 6 –> 3 –> 4 –> 5 –> 6, val = 6
Return: 1 –> 2 –> 3 –> 4 –> 5

java code

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

386ms =,=

ruby code实现如下：

# Definition for singly-linked list.
# class ListNode
#     attr_accessor :val, :next
#     def initialize(val)
#         @val = val
#         @next = nil
#     end
# end

# @param {ListNode} head
# @param {Integer} val
# @return {ListNode}
def remove_elements(head, val)
    while(head != nil && head.val == val)
        head = head.next
    end
    return nil if head == nil
    target = head
    while(target != nil && target.next != nil)
        if(target.next.val == val)
            target.next = target.next.next
        end
        if(target.next != nil && target.next.val != val)
            target = target.next
        end
    end
    return head
end

146ms 在这里解释执行还是快点。。。