代码随想录算法训练营第三天 | 203.移除链表元素 、707.设计链表 、 206.反转链表

203. 移除链表元素

链表好久不碰还真有点忘了，这题debug了大约15分钟，记录一下犯的错误：

1. cur.next如果要删除，cur在这个循环中不应该向后移动，否则会影响循环不变式，跳过某些要删的元素。

2. 返回应该返回dummy_head.next而不是head

# 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]:
        dummy_head = ListNode(0)
        dummy_head.next = head

        cur = dummy_head

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

707.设计链表 

这题一刷没做出来，二刷突然就开窍了...感觉这一年代码不是白写的...

没啥好说的，新建链表类实际新建虚拟头节点。

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

class MyLinkedList:

    def __init__(self):
        self.dummy_head = ListNode(0)

    def get(self, index: int) -> int:
        if index < 0:
            return -1

        cur = self.dummy_head

        for i in range(index + 1):
            if cur.next != None:
                cur = cur.next
            else:
                return -1

        return cur.val

    def addAtHead(self, val: int) -> None:
        tmp = ListNode(val)
        tmp.next = self.dummy_head.next
        self.dummy_head.next = tmp

    def addAtTail(self, val: int) -> None:
        cur = self.dummy_head
        tmp = ListNode(val)
        while cur.next != None:
            cur = cur.next
        cur.next = tmp

    def addAtIndex(self, index: int, val: int) -> None:
        if index < 0:
            return
        cur = self.dummy_head
        for i in range(index):
            if cur.next != None:
                cur = cur.next
            else:
                return
        tmp = ListNode(val)
        tmp.next = cur.next
        cur.next = tmp

    def deleteAtIndex(self, index: int) -> None:
        if index < 0:
            return
        cur = self.dummy_head
        for i in range(index):
            if cur.next != None:
                cur = cur.next
            else:
                return
        if cur.next != None:
            cur.next = cur.next.next
        


# 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

双指针实现比较简单，左指针指向反转好的新头节点，右指针指向还没翻转链表的的头节点，只需要保存右指针的next，将右指针指向的节点指到左指针指向的节点即可。

# 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]:
        slow, fast = None, head
        while fast != None:
            tmp = fast.next
            fast.next = slow
            slow = fast
            fast = tmp
        return slow

递归法比较难想明白，这里有头递归和尾递归两种写法，尾递归就是双指针；头递归则是递归返回时从后向前翻转。

头递归方法：

1. 终止条件：head是None，返回None， 处理空链表；head.next是None，这里给出新头节点。

2. 每一层递归调用栈中的new_head均是处理好的新链表的头节点，每层递归调用栈处于如下状态：

递归调用栈中的head → 本层栈中head → 翻转完成的链表尾节点 ← ... ← 新头节点

                            ↓

                            None

于是只需要让head.next.next指向head, head.next指向None即可返回到上一层递归调用栈。

# 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]:
        if head == None:
            return None
        if head.next == None:
            return head
        new_head = self.reverseList(head.next)
        head.next.next = head
        head.next = None
        return new_head