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

203.移除链表元素

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重点：使用虚拟头节点(dummynode)在删除或添加操作中更方便，能统一头节点操作写法

# 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)

        cur = dummy_head
        
        while cur.next:
            if cur.next.val == val:
                cur.next = cur.next.next
            else:
                cur = cur.next

        return dummy_head.next

---

707.设计链表

707.设计链表

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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

        cur = self.dummy_head.next

        while index:
            cur = cur.next
            index -= 1
        
        return cur.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:

        cur = self.dummy_head
        
        while cur.next:
            cur = cur.next
        cur.next = ListNode(val)
        self.size += 1


    def addAtIndex(self, index: int, val: int) -> None:

        if index < 0 or index > self.size:
            return 

        cur = self.dummy_head

        while index:
            cur = cur.next
            index -= 1

        # ListNode(val).next = cur.next
        # cur.next = ListNode
        cur.next = ListNode(val, cur.next)

        self.size += 1


    def deleteAtIndex(self, index: int) -> None:
        if index < 0 or index >= self.size:
            return 

        cur = self.dummy_head

        while index:
            cur = cur.next
            index -= 1

        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)

---

206.反转链表

206.反转链表

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重点：用双指针方法去做。初始化cur指向head, pre指向None。循环判断条件为什么是while cur:，当cur指向末尾(null)时循环结束，cur.next != null不行，它只能判断到倒数第二个元素，最后一个元素判断不了就结束了循环。

为什么先移动pre：如果先移动cur，cur指向temp，此时cur改变，pre无法指向原有的cur位置。

# 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