Leetcode[203] Remove Linked List Elements
16 Nov 2015###Task1 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
###Python ###Iterative
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def removeElements(self, head, val):
"""
:type head: ListNode
:type val: int
:rtype: ListNode
"""
dummy = ListNode(0)
dummy.next = head
ret = dummy
while dummy and dummy.next:
if dummy.next.val == val:
dummy.next = dummy.next.next
else:
dummy = dummy.next
return ret.next
###Recursive
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def removeElements(self, head, val):
"""
:type head: ListNode
:type val: int
:rtype: ListNode
"""
if not head:
return head
head.next = self.removeElements(head.next, val)
return head if head.val != val else head.next
###Java
/**
* 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) {
if (head == null) {
return null;
}
ListNode dummy = new ListNode(0);
dummy.next = head;
ListNode prev = dummy;
ListNode parser = head;
while (parser != null) {
if (parser.val == val) {
prev.next = parser.next;
} else {
prev = parser;
}
parser = parser.next;
}
return dummy.next;
}
}
###Points
Python采用递归法解题会返回Runtime Error，可能和栈溢出有关。 dummy node