【leetcode】链表 Linked List

发表于 更新于
链表的练习题（简单）

1、leetcode237：删除链表中的节点

请编写一个函数，使其可以删除某个链表中给定的（非末尾）节点，你将只被给定要求被删除的节点。

示例 1:
输入: head = [4,5,1,9], node = 5
输出: [4,1,9]
解释: 给定你链表中值为 5 的第二个节点，那么在调用了你的函数之后，该链表应变为 4 -> 1 -> 9.

示例 2:
输入: head = [4,5,1,9], node = 1
输出: [4,5,9]
解释: 给定你链表中值为 1 的第三个节点，那么在调用了你的函数之后，该链表应变为 4 -> 5 -> 9.

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# coding:utf-8
class Node:
    '''链表节点'''
    def __init__(self, x):
        self.val = x
        self.next = None

def listToLinkedList(list_):
    '''将列表转换成链表'''
    linkhead = Node(list_[0])
    p = linkhead
    # 逐个为列表内的数据创建结点, 建立链表
    for i in list_[1:]:
        p.next = Node(i)
        p = p.next
    return linkhead  # 链表头带着一串尾巴

def printLinkedList1(head):
    '''打印链表1'''
    if head == None:
        return False
    print(head.val)
    printLinkedList1(head.next)

def printLinkedList2(node):
    '''打印链表2(按照字符串打印)'''
    if not node:
        return "[]"
    result = ""
    while node:
        result += str(node.val) + ", "
        node = node.next
    print("[" + result[:-2] + "]")

#################################################

l = listToLinkedList([1,2,3,6,3,4,5])

def deleteNode(LinkedList, target):
    head = LinkedList
    p = head
    while p != None:
        if p.val == target:
            p.val, p.next = p.next.val, p.next.next
        else:
            p = p.next

deleteNode(l, 3)
printLinkedList2(l)

2、leetcode83：删除排序链表中的重复元素

给定一个排序链表，删除所有重复的元素，使得每个元素只出现一次。

示例 1:
输入: 1->1->2
输出: 1->2

示例 2:
输入: 1->1->2->3->3
输出: 1->2->3

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# coding:utf-8
class Node:
    '''链表节点'''
    def __init__(self, x):
        self.val = x
        self.next = None

def listToLinkedList(list_):
    '''将列表转换成链表'''
    linkhead = Node(list_[0])
    p = linkhead
    # 逐个为列表内的数据创建结点, 建立链表
    for i in list_[1:]:
        p.next = Node(i)
        p = p.next
    return linkhead  # 链表头带着一串尾巴

def printLinkedList1(head):
    '''打印链表1'''
    if head == None:
        return False
    print(head.val)
    printLinkedList1(head.next)

def printLinkedList2(node):
    '''打印链表2(按照字符串打印)'''
    if not node:
        return "[]"
    result = ""
    while node:
        result += str(node.val) + ", "
        node = node.next
    print("[" + result[:-2] + "]")

#################################################

l = listToLinkedList([1,2,2,3])

def deleteDuplicates(head):
    if head:
        before = head
        present = before.next
        while present:
            if present.val != before.val:
                before.next = present
                before = present
            else:
                before.next = None
            present = present.next
    return head

l1 = deleteDuplicates(l)
printLinkedList2(l1)

3、leetcode206：反转链表

反转一个单链表。

示例:
输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL

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class Node:
    '''链表节点'''
    def __init__(self, x):
        self.val = x
        self.next = None

def listToLinkedList(list_):
    '''将列表转换成链表'''
    linkhead = Node(list_[0])
    p = linkhead
    # 逐个为列表内的数据创建结点, 建立链表
    for i in list_[1:]:
        p.next = Node(i)
        p = p.next
    return linkhead  # 链表头带着一串尾巴

def printLinkedList1(head):
    '''打印链表1'''
    if head == None:
        return False
    print(head.val)
    printLinkedList1(head.next)

def printLinkedList2(node):
    '''打印链表2(按照字符串打印)'''
    if not node:
        return "[]"
    result = ""
    while node:
        result += str(node.val) + ", "
        node = node.next
    print("[" + result[:-2] + "]")

#################################################

l = listToLinkedList([1,2,3,4,5])

def reverseList(head):
    p, rev = head, None
    while p:
        rev, rev.next, p = p, rev, p.next
    return rev

l1 = reverseList(l)
printLinkedList2(l1)

4、leetcode21：合并两个有序链表

将两个有序链表合并为一个新的有序链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。

示例：
输入：1->2->4, 1->3->4
输出：1->1->2->3->4->4

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# coding:utf-8
class Node:
    '''链表节点'''
    def __init__(self, x):
        self.val = x
        self.next = None

def listToLinkedList(list_):
    '''将列表转换成链表'''
    linkhead = Node(list_[0])
    p = linkhead
    # 逐个为列表内的数据创建结点, 建立链表
    for i in list_[1:]:
        p.next = Node(i)
        p = p.next
    return linkhead  # 链表头带着一串尾巴

def printLinkedList1(head):
    '''打印链表1'''
    if head == None:
        return False
    print(head.val)
    printLinkedList1(head.next)

def printLinkedList2(node):
    '''打印链表2(按照字符串打印)'''
    if not node:
        return "[]"
    result = ""
    while node:
        result += str(node.val) + ", "
        node = node.next
    print("[" + result[:-2] + "]")

#################################################

l1 = listToLinkedList([1,2,3])
l2 = listToLinkedList([3,4,5])

def mergeTwoLists(l1, l2):
    if l1 and l2:
        if l1.val > l2.val: l1, l2 = l2, l1
        l1.next = mergeTwoLists(l1.next, l2)
    return l1 or l2

l3 = mergeTwoLists(l1, l2)
printLinkedList2(l3)