代码随想录算法训练营day3（链表）

移除链表元素

• https://leetcode.cn/problems/remove-linked-list-elements/

思路

• 问题：如果我们删除的元素是除头结点的其他元素，我们只需要将被删除元素的上一个节点的next指向被删除元素的下一个节点，但是如果删除的元素在头结点的话，我们就需要将head指针移动到下一个节点。这样就会导致整个操作不统一，需要判断被删除的节点是不是头结点
• 解决方案：采用虚拟头结点，这样整个操作过程就会统一

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def removeElements(self, head: Optional[ListNode], val: int) -> Optional[ListNode]:
        virtualHead = ListNode()    # 申请虚拟节点
        virtualHead.next = head    # 虚拟节点指向head
        cur = virtualHead           
        while cur.next:
            if cur.next.val == val:   
                cur.next = cur.next.next    # 删除节点
            else:
                cur = cur.next
        return virtualHead.next 

设计链表

• https://leetcode.cn/problems/design-linked-list/

• get(index)：获取链表中下标为index的值。
• addAtHead(val)：头部插入元素
• addAtTail(val)：尾部插入元素
• addAtIndex(index,val)：将节点插入到下标为index的位置
• deleteAtIndex(index)：删除下标为index的节点

思路

• 采用虚拟头结点

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

class MyLinkedList:
    def __init__(self):
        self.virtualHead = ListNode()
        self.size = 0

    def get(self, index: int) -> int:
        if index <0 or index > self.size -1:
            return -1
        cur = self.virtualHead.next
        for _ in range(index):
            cur = cur.next
        return cur.val

    def addAtHead(self, val: int) -> None:
        tempNode = ListNode(val)
        tempNode.next = self.virtualHead.next
        self.virtualHead.next = tempNode
        self.size +=1

    def addAtTail(self, val: int) -> None:
        tempNode = ListNode(val)
        cur = self.virtualHead    # 注意：没有节点时，尾部插入也可以是头部插入
        while cur.next:
            cur = cur.next
        cur.next = tempNode
        self.size +=1

    def addAtIndex(self, index: int, val: int) -> None:
        if index < 0 or index > self.size:   # 注意：最后一个节点的位置的下一个节点插入也是可以的
            return None
        tempNode = ListNode(val)
        cur = self.virtualHead       # 有可能在头结点的位置插入
        for _ in range(index):
            cur = cur.next
        tempNode.next = cur.next
        cur.next = tempNode
        self.size +=1

        
    def deleteAtIndex(self, index: int) -> None:
        if index <0 or index > self.size -1:
            return None
        cur = self.virtualHead    # 有可能删除头结点
        for _ in range(index):
            cur = cur.next
        cur.next = cur.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)

翻转链表

• https://leetcode.cn/problems/reverse-linked-list/

思路

• 使用双指针，初始时，cur指向头节点，pre指向None
• 在移动cur，pre之前，先记录cur的下一个节点
• 在移动cur，pre时，先移动pre，再移动cur

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        cur = head
        pre = None
        while cur:
            temp = cur.next
            cur.next = pre
            pre = cur
            cur = temp
        return pre