03天【代码随想录算法训练营34期】第二章 链表part01（203.移除链表元素 、707.设计链表 、 206.反转链表 ）

203.移除链表元素
竟然可以做个假head，学到了

class ListNode(object):
     def __init__(self, val=0, next=None):
         self.val = val
         self.next = next
class Solution(object):
    def removeElements(self, head, val):
        dummy_head = ListNode(-1)
        dummy_head.next = head

        current_node = dummy_head
        while current_node.next != None:
            if current_node.next.val == val:
                current_node.next = current_node.next.next
            else:
                current_node = current_node.next
        return dummy_head.next

707.设计链表
有点麻烦

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

class MyLinkedList(object):

    def __init__(self):
        self.head = None
        self.size = 0


    def get(self, index):
        if index < 0 or index >= self.size:
            return -1
        if self.head is None:
            return -1

        curr = self.head
        for i in range(index):
            curr = curr.next
        return curr.val
        
    def addAtHead(self, val):
        newNode = Node(val)
        newNode.next = self.head
        self.head = newNode
        self.size += 1
        
    def addAtTail(self, val):
        newNode = Node(val)
        curr = self.head
        if curr is None:
            self.head = newNode
        else:
            while curr.next:
                curr = curr.next
            curr.next = newNode
        self.size += 1
        
    def addAtIndex(self, index, val):
        if index < 0 or index > self.size:
            return
        if index == 0:
            self.addAtHead(val)
        else:
            newNode = Node(val)
            curr = self.head
            for i in range(index - 1):
                curr = curr.next
            newNode.next = curr.next
            curr.next = newNode
            self.size += 1

    def deleteAtIndex(self, index):
        if index < 0 or index >= self.size:
            return
        curr = self.head
        if index == 0:
            self.head = curr.next
        else:
            for i in range (index-1):
                curr = curr.next
            curr.next = curr.next.next
        
        self.size -= 1


# Your MyLinkedList object will be instantiated and called as such:
# obj = MyLinkedList()
# param_1 = obj.get(index)
# obj.addAtHead(val)
# obj.addAtTail(val)
# obj.addAtIndex(index,val)
# obj.deleteAtIndex(index)

206.反转链表

class ListNode(object):
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next
class Solution(object):
    def reverseList(self, head):
        tailNode = None
        newNode = None
        while head:
            newNode = ListNode(head.val, tailNode)
            tailNode = newNode
            head = head.next
        return newNode

If save space:

class Solution(object):
    def reverseList(self, head):
        # Initialize prev pointer as NULL...
        prev = None
        # Initialize the curr pointer as the head...
        curr = head
        # Run a loop till curr points to NULL...
        while curr:
            # Initialize next pointer as the next pointer of curr...
            next = curr.next
            # Now assign the prev pointer to curr’s next pointer.
            curr.next = prev
            # Assign curr to prev, next to curr...
            prev = curr
            curr = next
        return prev       # Return the prev pointer to get the reverse linked list...