2024-10-25 Leetcode刷题（二）

用户11029137

发布于 2024-10-31 08:12:04

6000

代码可运行

文章被收录于专栏：编程学习编程学习

运行总次数：0

代码可运行

LeetCode24

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def swapPairs(self, head: Optional[ListNode]) -> Optional[ListNode]:
        node0 = dummy = ListNode(next=head)
        node1 = head
        while node1 and node1.next:
            node2 = node1.next
            node3 = node2.next

            node0.next = node2
            node2.next = node1
            node1.next = node3

            node0 = node1
            node1 = node3
        return dummy.next

在有可能对头结点进行操作的时候，创建哨兵节点简化

注意：head本身指向链表第一个元素，在代码中head未被改变，运行后，head不一定指向第一个元素而dummy是虚拟头结点，在第一次循环中dummy.next指向了第一个元素，而后续，并未改变虚拟头结点的指向，所以dummy.next始终指向的是第一个元素

Leetcode160

（默认相交以后的都相同？）

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

class Solution:
    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> Optional[ListNode]:
        A, B = headA, headB
        while A != B:
            A = A.next if A else headB
            B = B.next if B else headA
        return A

1，分别遍历两个链表，指针的总路程是相同的 2，第一遍遍历是已走路程相同，第二次遍历是剩余路程相同 3，把NULL当做两个链表一定有的共同点！

Leetcode 203

# 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]:
        cur = dummy = ListNode(next=head)
        while cur.next:
            if cur.next.val == val:
                cur.next = cur.next.next
            else:
                cur = cur.next
        return dummy.next

LeetCode 19

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
        left = right = dummy = ListNode(next=head)
        for _ in range(n):
            right = right.next
        while right.next:
            left = left.next
            right = right.next
        
        left.next = left.next.next
        return dummy.next

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原始发表：2024-10-30，如有侵权请联系 cloudcommunity@tencent.com 删除

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