python 链表

在C/C++中，通常采用“指针+结构体”来实现链表；而在Python中，则可以采用“引用+类”来实现链表。

节点类：

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

链表类：

class Linkedlist:
    def __init__(self):
        self.head = None
        self.tail = None
link_list = LinkedList()

def is_empty(self):
    return self.head is None

def append(self, data):
    node = Node(data)
    if self.head is None:
        self.head = node
        self.tail = node
    else:
        self.tail.next = node
        self.tail =node

def iter(self):
    if not iter.head:
        return
    cur = self.head
　　 yield cur.data
while cur.next:
    cur = cur.next
    yield cur.data
#先判断是不是空链表，yield head.data 再用while循环遍历

链表的头结点head 和 尾节点tail 都属于node.

insert：先将要插入的节点的next指向之后链表的head，然后将之前链表的next指向 将要插入的节点。

def insert(self, idx, value):
    cur = self.head
    cur_idx = 0
    if cur is None:
        raise Exception('That list is and empty list!')
    while cur_idx < idx-1:
        cur = cur.next
        if cur is None:
            raise Exception('List length less than index!')
        cur_idx += 1
    node = Node(value)
    node.next = cur.next
    cur.next = node
    if node.next is None:
        self.tail = node

def remove(self, idx):
    cur = self.head
    cur_idx = 0
    #空指针
    if self.head = None:
        raise Exception('This is an empty list')
    while cur_idx < idx-1:
        cur = cur.next
        #给出的索引大于链表的长度
        if cur is None:
            raise Exception('list length less than index')
        cur_idx +=1
    if idx == 0:    #当删除第一个节点时
        self.head = cur.next
        cur = cur.next
        return
    if self.head is self.tail: #当只有一个节点时
        self.head = None
        self.tail = None
        return
    cur.next = cur.next.next
    if cur.next is None:    #当删除最后一个节点时
        self.tail = cur        

def size(self):
    i = 0
    cur = self.head
    if current is None:
        return 'The list is an empty list'
    while cur.next is not None:
        i +=1
        cur = cur.next
    return i

def search(self, item):
    current = self.head
    found = False
    while current is not None and not found:
        if current.data == item:
            found = True
        else:
            current = current.next
    return found

单链表逆置

1，迭代

# -*- coding: utf-8 -*-
#!/bin/env python
# Python2.7class Node(object):
    def __init__(self):
        self.value = None
        self.next = None
    def __str__(self):
        return str(self.value)def reverse_list(head):
    if not head or not head.next:
        return head
    pre = None
    while head:
        next = head.next    # 缓存当前节点的向后指针，待下次迭代用
        head.next = pre     # 关键:把当前节点向前指针(pre)作为当前节点的向后指针
        pre = head          # 把当前指针赋值给 下次迭代 节点的 向前指针
        head = next         # 作为下次迭代时的（当前）节点
    return pre    # 返回头指针，头指针就是迭代最后一次的head（赋值给类pre）if __name__ == '__main__':    three = Node()
    three.value = 3    two = Node()
    two.value = 2
    two.next = three    one = Node()
    one.value = 1
    one.next = two    head = Node()
    head.value = 0
    head.next = one    newhead = reverse_list(head)
    while newhead:
        print newhead.value
        newhead = newhead.next

比较形象的图

2，递归

# 临界点：head.next为None
# 先递归到 把最后一个节点指向 newhead
# 然后一步步从后往前逆置def reverse_recursion(head):
    if not head or not head.next:
        return head    new_head = reverse_recursion(head.next)    head.next.next = head
    head.next = None
    return new_head