LeetCode解题方法之——空指针_listnode怎么创造空指针

203. Remove Linked List Elements

1.创建空的头指针

1.这样可以处理空链表特殊情况，头指针为要删除的值的特殊情况
使用空指针p在这里的好处在于，
1.p保存了头结点，
2.而最开始pre和p指向了同一个节点对象，这样如果首节点是空节点或等于val时，可以直接进行操作，因此首节点也可以当做普通节点来处理，返回值是p.next即可，这种解法的巧妙之处就在这里。

class Solution:
    def removeElements(self, head, val):
        dummyNode = ListNode(0)
        dummyNode.next = head
        pre = dummyNode
        cur = head
        while cur:
            if cur.val==val:
                pre.next = cur.next
            else:
                pre = cur
            cur = cur.next
        return dummyNode.next
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2.循环解法

class Solution:
    def removeElements(self, head, val):
        
        if not head:
            return None
        while head!=None and head.val==val:
            head = head.next
        pre = None
        cur = head
        while cur:
            if cur.val==val:
                pre.next = cur.next
            else:
                pre = cur
            cur = cur.next
        return head
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class Solution:
    def removeElements(self, head, val):
        """
        :type head: ListNode
        :type val: int
        :rtype: ListNode
        """
        while head and head.val==val:
            head=head.next
        p = head
        while p and p.next:
            if p.next.val == val:
                p.next = p.next.next
            else:
                p = p.next
        return head
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3. 递归解法

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

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