在O(1)时间删除链表节点

在O(1)时间删除链表节点

【题目】：

给定单向链表的头指针和一个结点指针，定义一个函数在O(1)时间删除该节点。

void deleteNode(ListNode* pListHead, ListNode* pToBeDeleted)

【解题思路】：

从三个方面来考虑：

• 如果待删除节点不是尾结点：将该节点的值由下一个节点的值来代替，然后指向下下个节点；
• 如果待删除节点是尾结点同时也是头节点，链表中只有一个节点： 将头节点置为NULL；
• 如果是尾巴节点，则需要遍历到尾节点前的一个节点，然后进行删除操作。  

#  defined a linklist
class NodeList(object):
    def __init__(self, val):
        self.val = val
        self.next = None


#  creat a linklist
def creat_linklist(list):
    if not list:
        return
    head = p = NodeList(list[0])
    for i in list[1:]:
        p.next = NodeList(i)
        p = p.next
    return head


#  print a linklist
def print_linklist(head):
    if not head:
        print("This is an empty linklist!")
        return
    while head:
        print(head.val, end='->')
        head = head.next
    print(None)


#  delete a node
def delete_a_node(head, node):
    if not head and not node:
        return
    if node.next != None:
        node.val = node.next.val
        node.next = node.next.next
    elif head == node:
        head = None
    else:
        p = head
        while p.next != node:
            p = p.next
        p.next = None
    return head


# test
# 1. create a linklist
head = creat_linklist([1, 2, 3, 4, 5])
print_linklist(head)
# 2. delete a node in the middle of linklist
toBeDeleted = head.next.next
new_head = delete_a_node(head, toBeDeleted)
print_linklist(new_head)
# 3. delete a node in bottom of the linklist
toBeDeleted = head.next.next.next
new_head = delete_a_node(head, toBeDeleted)
print_linklist(new_head)
# 4. test a one node linklist
head = creat_linklist([1])
new_head = delete_a_node(head, head)
print_linklist(new_head)

运行结果：