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

Leetcode 203 Remove Linked List Elements

• 题目链接：leetcode.com/problems/re…
• 文章讲解：代码随想录 programmercarl.com
• 视频讲解：手把手带你学会操作链表 | LeetCode：203.移除链表元素
• 状态：没做出来

1. 第一想法

我忘记怎么遍历 linked list 了。

2. 看完后想法

提到两种做法，一种是原链表上操作，另一种是使用虚拟头节点。前者的缺点是不够统一，要分别考虑删头节点和非头节点的情况。这里只贴了后者的代码。

# 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(next = head)
        
        current = dummy_head
        while current.next:
            if current.next.val == val:
                current.next = current.next.next
            else:
                current = current.next
        return dummy_head.next

一开始我最后return head其实过了例子诶，但是提交一个全要删得没过。应该是因为那个例子只要删最后一个节点所以运气通过了。

3. 总结

其实以前一直不知道为啥以及什么情况要 dummy node，一直是试着加一下，没过删一下。这下明白为啥了。

Leetcode 707 Design Linked List

• 题目链接：leetcode.com/problems/de…
• 文章讲解：代码随想录 programmercarl.com
• 视频讲解：帮你把链表操作学个通透！LeetCode：707.设计链表
• 状态：没做出来

1. 第一想法

每次遇到什么 design 或者 build 我都没想法。

2. 看完后想法

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

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

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

    def addAtTail(self, val: int) -> None:
        current = self.dummy_head
        while current.next:
            current = current.next
        current.next = ListNode(val)
        self.size += 1

    def addAtIndex(self, index: int, val: int) -> None:
        if index < 0 or index > self.size:
            return
        current = self.dummy_head
        for i in range(index):
            current = current.next
        current.next = ListNode(val, current.next)
        self.size += 1

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

3. 总结

很全面的一题，常看常新。

Leetcode 206 Reverse Linked List

• 题目链接：leetcode.com/problems/re…
• 文章讲解：代码随想录 programmercarl.com
• 视频讲解：帮你拿下反转链表 | LeetCode：206.反转链表 | 双指针法 | 递归法
• 状态：没做出来

1. 第一想法

觉得应该让指针顺序倒过来....但是不知道怎么处理指向上一个。

2. 看完后想法

天呐，我怎么就没想到双指针！

# 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
        prev = None
        while cur:
            temp = cur.next
            cur.next = prev
            prev = cur
            cur = temp
        return prev

递归其实和双指针差不多。我就是那种依葫芦画瓢凭感觉写题第二次遇到又不会的人...

# 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]:
        return self.reverse(head, None)
    
    def reverse(self, cur: ListNode, prev: ListNode) -> ListNode:
        if cur == None:
            return prev
        temp = cur.next
        cur.next = prev
        return self.reverse(temp, cur)

3. 总结

主要还是要掌握双指针的思路。