203. 移除鏈表元素 - 力扣（LeetCode）

之前沒(méi)接觸過(guò)鏈表盲憎，直接看講解

解題思路

有兩種嗅骄，單獨(dú)處理頭節(jié)點(diǎn)，或者設(shè)置虛擬頭節(jié)點(diǎn)饼疙，讓頭節(jié)點(diǎn)作為普通節(jié)點(diǎn)

方法一 使用原鏈表

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next

class Solution(object):
    def removeElements(self, head, val):
        """
        :type head: ListNode
        :type val: int
        :rtype: ListNode
        """
        while head :
            if head.val == val:
                head = head.next
            else:
                break

        current = head

        while current and current.next is not None:
            if current.next.val == val:
                current.next = current.next.next
            else:
                current = current.next

        return head

方法二 虛擬頭節(jié)點(diǎn)

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next

class Solution(object):
    def removeElements(self, head, val):
        """
        :type head: ListNode
        :type val: int
        :rtype: ListNode
        """
        dummy_head = ListNode()
        dummy_head.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

707. 設(shè)計(jì)鏈表 - 力扣（LeetCode）

單鏈表

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

class MyLinkedList(object):

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

    def get(self, index):
        """
        :type index: int
        :rtype: 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):
        """
        :type val: int
        :rtype: None
        """
        self.dummy_head.next = ListNode(val, self.dummy_head.next)
        self.size += 1


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

    def addAtIndex(self, index, val):
        """
        :type index: int
        :type val: int
        :rtype: 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):
        """
        :type index: int
        :rtype: 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

• 下標(biāo)是從dummy_head.next開(kāi)始溺森，從0開(kāi)始
• 參考代碼隨想錄 (programmercarl.com)

206. 反轉(zhuǎn)鏈表 - 力扣（LeetCode）

雙指針?lè)?/h2>

    def reverseList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """

        cur = head
        pre = None

        while cur:
            temp = cur.next #保存cur下一個(gè)節(jié)點(diǎn)
            cur.next = pre # 反轉(zhuǎn)

            pre = cur  #面向?qū)ο?            cur = temp           

        return pre

遞歸

class Solution(object):
    def reverseList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        return self.reverseNode(head, None)
    
    def reverseNode(self, cur, pre):
        if cur is None:
            return pre

        temp = cur.next
        cur.next = pre
        return self.reverseNode(temp, cur)
               

• 注意類的寫(xiě)法
• 遞歸還需練習(xí)