数据结构与算法7 - 链表

链表：(链表应用：操作系统的缓存、Dictionary)

　　1. 链表由一系列数据记录构成，在每个记录里有个区域包含一个指向下一个数据记录的索引

　　2. 链表vs数组列表

　　　　（1）使用固定步数的操作可以在列表中任意节点做插入、删除操作

　　　　（2）不允许随机访问

　　3.　循环链表和双向链表

　　　　

 　　4. 哨兵节点

　　　　在某些实现中，可能会在第一个数据记录之前或者最后一个数据记录之后添加一个额外的哨兵节点或者哑元节点

　　　　简化和加快一些列表处理的算法

　　　　链表vs.动态数组
　　　　　　— 都需要动态分配内存块
　　　　— 动态数组：
　　　　　　— 通过索引访问和分配非常快速，时间复杂度为O(1)
　　　　　　— 添加元素（插入到数组的末尾）相对比较快，平摊时间复杂度为O(1)
　　　　　　— 在动态数组里的任意位置添加和删除节点会很慢，时间复杂度为O(n)
　　　　　　— 当需要对插入和删除做调整时，可能会出现不可预知的表现。
　　　　　　— 会有一些未使用的空间
　　　　— 链表：
　　　　　　— 在列表中的任意位置做插入和删除都很快，时间复杂度为O(1)
　　　　　　— 索引访问(随机访问)慢，时间复杂度为O(n)

　　5. 抽象数据类型（ADT）列表操作

　　　　（1）创建一个空列表

　　　　（2）判断列表是否为空

　　　　（3）确定列表中元素个数

　　　　（4）在列表中给定位置添加一个元素

　　　　（5）在列表中给定位置删除一个元素

　　　　（6）删除列表中所有元素

　　　　（7）在列表中取到给定位置上的元素

　　　　（8）每一项操作的时间复杂度

class Node:
    def __init__(self, value = None, next = None):
        self.value = value
        self.next = next
class LinkedList:
    def __init__(self):
        self.head = Node(None, None)
        self.size = 0

    def get_first(self):
        if not self.head.next:
            print('LinkedList is empty')
        return self.head.next
    def get_last(self):
        if not self.head.next:
            print('LinkedList is empty')
        node = self.head
        while node.next != None:
            node = node.next
        return node

    def get(self, index):
        if (index < 0 or index >= self.length):
            print('index is out of bound');
        if not self.head.next:
            print('LinkedList is empty')
        node = self.head.next
        for i in range(index):
            node = node.next
        return node
    def add_first(self, value):
        node = Node(value, None)
        node.next = self.head.next
        self.head.next = node
        self.size += 1
    def add_last(self, value):
        new_node = Node(value,None)
        node = self.head
        while node.next != None:
            node = node.next
        node.next = new_node
        self.size += 1
    def remove_first(self):
        if not self.head.next:
            print("the linked list is empty")
        value = self.head.next
        self.head.next = self.head.next.next
        self.size -= 1
        return value
    def remove_last(self):
        if not self.head.next:
            print("the linked list is empty")
        node = self.head.next
        pre = self.head
        while node.next != None:
            pre = node
            node = node.next
        pre.next = None
        self.size -= 1
        return node.value
    def add(self, index, value):
        if (index < 0 or index > self.size):
            print('index is out of bound')
        if not self.head.next:
            print('LinkedList is empty')
        new_node = Node(value)
        node = self.head
        for i in range(index):
            node = node.next
        new_node.next = node.next;
        node.next = new_node;
        self.size += 1
    def remove(self, index):
        if (index < 0 or index >= self.size):
            print('index is out of bound');
        if not self.head.next:
            print("the linked list is empty")
        node = self.head
        for i in range(index):
            node = node.next
        result = node.next
        node.next = node.next.next
        self.size -= 1
        return result
    def print_list(self):
        node = self.head
        while node.next != None:
            node = node.next
            print(node.value, end = " ")
        print()
    def length(self):
        return self.size

ll = LinkedList()
ll.add_first(5)
ll.add(1,8)
ll.add_last(14)
print(ll.length())
ll.print_list()

　　6. 练习1：删除链表中的节点，除了结尾，只允许访问那个节点

def delete_node(node):
    print(node.value)
    node.value = node.next.value
    node.next = node.next.next

　　7. 练习2：找到中间节点

def find_middle(lst):
    assert lst.head is not None and lst.head.next is not None
    
    head = lst.head
    fast = head
    slow = head
    
    while fast is not None and fast.next is not None:
        fast = fast.next.next
        slow = slow.next
    
    return slow.value

　　8. 练习3：是否有环

def has_cycle(lst):
    return has_cycle_helper(lst.head)

def has_cycle_helper(head):
    if head is None:
        return False
    
    slow = head 
    fast = head
    
    while fast is not None and fast.next is not None:
        fast = fast.next.next
        slow = slow.next
        
        if slow==fast:
            return True
        
    return False

　　9. 练习4：找到环的开始：先跑两遍，一个快的是两倍速一个慢是一倍速，两者相遇后，将快的重归于原点，慢的继续跑，两个再同样速度跑，相遇就是环的起点

def find_beginning(head):
    if head is None:
        return None
    
    slow = head
    fast = head
    
    while fast is not None and fast.next is not None:
        fast = fast.next.next
        slow = slow.next
        
        if slow==fast:
            fast = head
            break
        
    if fast is None or fast.next is None:
        return None
    
    while fast != slow:
        fast = fast.next
        slow = slow.next

    return slow

　　9. 练习5：删除倒数第N个节点（限制：不能用size）

def remove_nth(lst, n):
    assert n<=lst.length and n > 0
    
    fast = lst.head
    while n>0:
        fast = fast.next
        n = n - 1
        
    slow = lst.head
    while fast.next is not None:
        fast = fast.next
        slow = slow.next
        
    result = slow.next
    slow.next = slow.next.next
    
    lst.length = lst.length - 1
        
    return result

　　10. 练习6：分半：给定一个列表，把它分成两个列表，一个是前半部分，一个是后半部分

def split(head):
    if (head is None):
        return
    slow = head
    fast = head
    front_last_node = slow
    while (fast is not None):
        front_last_node = slow
        slow = slow.next
        fast = fast.next.next if fast.next is not None else None
    front_last_node.next = None
    front = head
    back = slow
    return (front, back)

 　　11. 练习7：合并两个排序链表

　　Input: 1->2->4, 1->3->4

　　Output: 1->1->2->3->4->4

def mergeTwoLists1(l1, l2):
    dummy = cur = Node(0)
    while l1 and l2:
        if l1.value < l2.value:
            cur.next = l1
            l1 = l1.next
        else:
            cur.next = l2
            l2 = l2.next
        cur = cur.next
    cur.next = l1 or l2
    return dummy.next

def mergeTwoLists2(l1, l2):
    if not l1 or not l2:
        return l1 or l2
    if l1.value < l2.value:
        l1.next = mergeTwoLists2(l1.next, l2)
        return l1
    else:
        l2.next = mergeTwoLists2(l1, l2.next)
        return l2

　　练习8：

　　　找到两个相同链表的起始位置

　　　　A: a1 → a2

             　　　　  ↘

               　　　　　　  c1 → c2 → c3

              　　　　 ↗    

　　　　B: b1 → b2 → b3

　　. 题解：先让AB同时跑，A先到终点，然后让A从B开始的位置再跑，B到终点后从A开始的位置继续跑，两者到达C1相遇

def getIntersectionNode2(headA, headB):
    if headA and headB:
        A, B = headA, headB
        while A!=B:
            A = A.next if A else headB
            B = B.next if B else headA
        return A

　　练习9：排序一个链表O（nlgn）

　　先找到中点，然后将一个链表拆分为两个链表，然后将每一个链表排序合并

def sortList(head):
    if head is None or head.next is None:
        return head
    mid = getMiddle(head)
    rHead = mid.next
    mid.next = None
    return merge(sortList(head), sortList(rHead))

def merge(lHead, rHead):
    dummyNode = dummyHead = Node(0)
    while lHead and rHead:
        if lHead.value < rHead.value:
            dummyHead.next = lHead
            lHead = lHead.next
        else:
            dummyHead.next = rHead
            rHead = rHead.next
        dummyHead = dummyHead.next
    if lHead:
        dummyHead.next = lHead
    elif rHead:
        dummyHead.next = rHead
    return dummyNode.next

def getMiddle(head):
    if head is None:
        return head
    slow = head
    fast = head
    while fast.next and fast.next.next:
        slow = slow.next
        fast = fast.next.next
    return slow

　　练习10：Partition 链表

　　给一个变量x, 链表左边所有的元素都比x小，x值右边所有元素值都比x大

def partition(head, x):
    left_head = Node(None)  # head of the list with nodes values < x
    right_head = Node(None)  # head of the list with nodes values >= x
    left = left_head  # attach here nodes with values < x
    right = right_head  # attach here nodes with values >= x
    # traverse the list and attach current node to left or right nodes
    while head:
        if head.value < x:
            left.next = head
            left = left.next
        else:  # head.val >= x
            right.next = head
            right = right.next
        head = head.next
    right.next = None  # set tail of the right list to None
    left.next = right_head.next  # attach left list to the right
    return left_head.next  # head of a new partitioned list

　　 练习11：反转一个链表

def reverse(lst):
    head = lst.head
    result = None
    current = head.next
    nxt = None
    
    while current is not None:
        nxt = current.next
        current.next = result
        result = current
        current = nxt
        
    head.next = result

def reverseRecursion(node):
    if (node is None or node.next is None):
        return node
    p = reverseRecursion(node.next)
    node.next.next = node
    node.next = None
    return p