LeetCode - Linked List 总结

最新推荐文章于 2024-04-19 16:41:16 发布

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这篇博客总结了LeetCode中涉及链表的几道经典题目，包括：加两个数、合并两个有序链表、合并k个有序链表、反转链表、检测链表环、删除链表倒数第n个节点以及判断回文链表等。

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2. Add Two Numbers

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

class Solution:
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        head = ListNode(0)
        cur = head
        carry = 0
        while l1 or l2 or carry:
            if l1:
                carry += l1.val
                l1 = l1.next
            if l2:
                carry += l2.val
                l2 = l2.next
            cur.next = ListNode(carry%10)
            carry = int(carry/10)
            cur = cur.next
        return head.next

21. Merge Two Sorted Lists

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

class Solution:
    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
        head = ListNode(0)
        cur = head
        while l1 and l2:
            if l1.val < l2.val:
                cur.next = l1
                l1 = l1.next
            else:
                cur.next = l2
                l2 = l2.next
            cur = cur.next
        if l1:
            cur.next = l1
        if l2:
            cur.next = l2
        return head.next

23. Merge k Sorted Lists

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

class Solution(object):
    def mergeKLists(self, lists):
        """
        :type lists: List[ListNode]
        :rtype: ListNode
        """
        head = cur = ListNode(0)
        lists = [(i.val, i) for i in lists if i]
        heapq.heapify(lists)
        while lists:
            cur.next = heapq.heappop(lists)[1]
            cur = cur.next
            if cur.next:
                heapq.heappush(lists, (cur.next.val, cur.next))
        return head.next

206. Reverse Linked List

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

class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        pre = None
        cur = head
        while cur:
            nextNode = cur.next
            cur.next = pre
            pre = cur
            cur = nextNode
        return pre

141. Linked List Cycle

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

class Solution(object):
    def hasCycle(self, head):
        """
        :type head: ListNode
        :rtype: bool
        """
        slow = head
        fast = head
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
            if fast==slow: return True
        return False

19. Remove Nth Node From End of List

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

class Solution(object):
    def removeNthFromEnd(self, head, n):
        """
        :type head: ListNode
        :type n: int
        :rtype: ListNode
        """
        dummy = ListNode(0)
        dummy.next = head
        slow = dummy
        fast = dummy
        for i in range(0, n+1):
            fast = fast.next
        while fast:
            slow = slow.next
            fast = fast.next
        slow.next = slow.next.next
        return dummy.next

234. Palindrome Linked List

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

class Solution(object):
    def isPalindrome(self, head):
        """
        :type head: ListNode
        :rtype: bool
        """
        if head is None: return True
        slow = fast = head
        while fast.next and fast.next.next:
            slow = slow.next
            fast = fast.next.next
        pre, cur = None, slow.next
        while cur:
            nextNode = cur.next
            cur.next = pre
            pre, cur = cur, nextNode
        
        while pre:
            if head.val!=pre.val:
                return False
            head = head.next
            pre = pre.next
        return True

160. Intersection of Two Linked Lists

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

class Solution(object):
    def getIntersectionNode(self, headA, headB):
        """
        :type head1, head1: ListNode
        :rtype: ListNode
        """
        if not headA or not headB: return None
        # get the end of list A
        endA = headA
        while endA.next:
            endA = endA.next
        endA.next = headB
        
        # get the intersection
        slow = headA
        fast = headA
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
            if slow==fast: break
            
        if slow!=fast: 
            endA.next = None
            return None
        
        fast = headA
        while fast:
            if fast==slow: 
                endA.next = None
                return fast
            fast = fast.next
            slow = slow.next