二刷 leetcode （9）

最新推荐文章于 2025-03-21 17:45:22 发布

原创最新推荐文章于 2025-03-21 17:45:22 发布 · 332 阅读

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#环形链表 #合并有序链表 #回文链表

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本文介绍如何合并两个有序链表、检测链表中的环以及判断链表是否为回文链表。提供了多种实现方法，包括使用快慢指针检测环、递归与非递归方式合并链表等。

21. 合并两个有序链表

将两个有序链表合并为一个新的有序链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。

示例：

输入：1->2->4, 1->3->4
输出：1->1->2->3->4->4

代码：

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

class Solution:
    def mergeTwoLists(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        ####  the first method
        newHead = ListNode(0)
        pre = newHead
        while l1 and l2:
            if l1.val < l2.val:
                newHead.next = l1
                l1 = l1.next
            else:
                newHead.next = l2
                l2 = l2.next
            newHead = newHead.next
        if l1:
            newHead.next = l1
            list.append(pre.next.val)      
        elif l2:
            newHead.next = l2
            list.append(pre.next.val)
        return pre.next
        
        #### the second method
        if l1 == None:
            return l2
        if l2 == None:
            return l1
        dummy = ListNode(0)
        tmp = dummy
        while l1 and l2:
            if l1.val <= l2.val:
                tmp.next = l1
                l1 = l1.next
                tmp = tmp.next
            else:
                tmp.next = l2
                l2 = l2.next
                tmp = tmp.next
        if l2 == None:
            tmp.next = l1
        else:
            tmp.next = l2
        return dummy.next

141. 环形链表

给定一个链表，判断链表中是否有环。

代码;

# 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
        """
        #### the first method 
        
        # slow = fast = head
        # if head == None or head.next == None:
        #     return False
        # while fast and fast.next:
        #     slow = slow.next
        #     fast = fast.next.next
        #     if fast == slow:
        #         return True
        # return False
        
        #### the second method
        
        slow = fast = head
        while fast:
            slow, fast = slow.next, fast.next
            if not fast:
                return False
            fast = fast.next
            if fast == slow:
                return True
        return False

234. 回文链表

给定一个链表，判断链表中是否有环。

请判断一个链表是否为回文链表。

示例 1:

输入: 1->2
输出: false

示例 2:

输入: 1->2->2->1
输出: true

代码:

# 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
        """
        #####  the first method
#         if head == None or head.next == None:
#             return True
#         values = []
#         p = head
#         while p.next:
#             values.append(p.val)
#             p = p.next
#         values.append(p.val)
        
#         return values[:] == values[::-1]

        ##### the second method
        if head is None or head.next is None:
            return True
        mid = ListNode(0)
        mid.next = head
        fast = mid
        while fast and fast.next:
            mid = mid.next
            fast = fast.next.next
        
        pre = mid 
        cur = mid.next
        while cur:
            temp  = cur.next
            cur.next = pre
            pre = cur
            cur = temp
        
        start = head
        end = pre
        if start.val != end.val:
            return False
        while start != mid.next:
            if start.val != end.val:
                return False
            start = start.next
            end = end.next
        return True

给定一个链表，判断链表中是否有环。