LeetCode Hot100 刷题笔记（3）—

LeetCode Hot100 刷题笔记（3）—— 链表

news2025/4/8 3:29:21

目录

前言

1. 相交链表

2. 反转链表

3. 回文链表

4. 环形链表

5. 环形链表 II

6. 合并两个有序链表

7. 两数相加

8. 删除链表的倒数第 N 个结点

9. 两两交换链表中的节点

10. K 个一组翻转链表

11. 随机链表的复制

12. 排序链表

13. 合并 K 个升序链表

14. LRU 缓存

前言

一、链表：相交链表，反转链表，回文链表，环形链表，环形链表 II，合并两个有序链表，两数相加，删除链表的倒数第 N 个结点，两两交换链表中的节点，K 个一组翻转链表，随机链表的复制，排序链表，合并 K 个升序链表，LRU 缓存。(日更中...)

*** Trick：本质将链表转为list，再在list上进行操作，最后转回链表。

*** Trick 通用模版：

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

class Solution(object):
    def Operation(self, head):
        if not head:
            return None
        lst = []
        while head:
            lst.append(head.val)
            head = head.next
        """
        系列操作
        """
        head_new = ListNode(int(lst[0]))
        curr = head_new
        for v in lst[1:]:
            curr.next = ListNode(int(v))
            curr = curr.next
        return head_new

1. 相交链表

原题链接：160. 相交链表 - 力扣（LeetCode）

class Solution(object):
    def getIntersectionNode(self, headA, headB):
        set1 = set()
        while headA:
            set1.add(headA)
            headA = headA.next
        while headB:
            if headB in set1:
                return headB
            headB = headB.next
        return None

2. 反转链表

原题链接：206. 反转链表 - 力扣（LeetCode）

class Solution(object):
    def reverseList(self, head):
        if not head:
            return None
        lst = []
        while head:
            lst.append(head.val)
            head = head.next
        lst.reverse()

        head_new = ListNode(int(lst[0]))
        curr = head_new
        for v in lst[1:]:
            curr.next = ListNode(int(v))
            curr = curr.next
        return head_new

3. 回文链表

原题链接：234. 回文链表 - 力扣（LeetCode）

class Solution(object):
    def isPalindrome(self, head):
        if not head:
            return None
        lst = []
        while head:
            lst.append(head.val)
            head = head.next
        if lst == lst[::-1]:
            return True
        return False

4. 环形链表

原题链接：141. 环形链表 - 力扣（LeetCode）

class Solution(object):
    def hasCycle(self, head):
        if not head:
            return False
        set1 = set()
        while head:
            set1.add(head)
            head = head.next
            if head in set1:
                return True
        return False

5. 环形链表 II

原题链接：142. 环形链表 II - 力扣（LeetCode）

class Solution(object):
    def detectCycle(self, head):
        if not head:
            return None
        set1 = set()
        while head:
            set1.add(head)
            head = head.next
            if head in set1:
                return head
        return None

6. 合并两个有序链表

原题链接：21. 合并两个有序链表 - 力扣（LeetCode）

class Solution(object):
    def mergeTwoLists(self, list1, list2):
        if not list1 and not list2:
            return None
        elif not list1:
            return list2
        elif not list2:
            return list1
        else:
            lst1, lst2 = [], []
            while list1:
                lst1.append(list1.val)
                list1 = list1.next
            while list2:
                lst2.append(list2.val)
                list2 = list2.next
            lst = lst1 + lst2
            lst.sort()

            head = ListNode(int(lst[0]))
            curr = head
            for v in lst[1:]:
                curr.next = ListNode(int(v))
                curr = curr.next
            return head

7. 两数相加

原题链接：2. 两数相加 - 力扣（LeetCode）

class Solution(object):
    def addTwoNumbers(self, l1, l2):
        lst1, lst2 = [], []
        while l1:
            lst1.append(l1.val)
            l1 = l1.next
        while l2:
            lst2.append(l2.val)
            l2 = l2.next
        # s1 = ''.join(l1)
        # s2 = ''.join(l2)
        lst1.reverse()
        lst2.reverse()
        s1 = ''.join([str(i) for i in lst1])
        s2 = ''.join([str(i) for i in lst2])
        s3 = int(s1) + int(s2)
        lst3 = list(str(s3))
        lst3.reverse()

        head = ListNode(int(lst3[0]))
        curr = head
        for v in lst3[1:]:
            curr.next = ListNode(int(v))
            curr = curr.next
        return head

8. 删除链表的倒数第 N 个结点

原题链接：19. 删除链表的倒数第 N 个结点 - 力扣（LeetCode）

class Solution(object):
    def removeNthFromEnd(self, head, n):
        lst = []
        while head:
            lst.append(head.val)
            head = head.next
        del lst[-n]
        if not lst:
            return None
        else:
            head_new = ListNode(lst[0])
            curr = head_new
            for v in lst[1:]:
                curr.next = ListNode(v)
                curr = curr.next
            return head_new

9. 两两交换链表中的节点

原题链接：24. 两两交换链表中的节点 - 力扣（LeetCode）

class Solution(object):
    def swapPairs(self, head):
        if not head:
            return None
        lst = []
        while head:
            lst.append(head.val)
            head = head.next
        for i in range(0, len(lst)-1, 2):
            lst[i], lst[i+1] = lst[i+1], lst[i]
        head_new = ListNode(lst[0])
        curr = head_new
        for value in lst[1:]:
            curr.next = ListNode(value)
            curr = curr.next
        return head_new

10. K 个一组翻转链表

原题链接：25. K 个一组翻转链表 - 力扣（LeetCode）

class Solution(object):
    def reverseKGroup(self, head, k):
        lst = []
        while head:
            lst.append(head.val)
            head = head.next
        for i in range(0, len(lst)-k+1, k):
            lst[i:i+k] = lst[i:i+k][::-1]

        head_new = ListNode(lst[0])
        curr = head_new
        for v in lst[1:]:
            curr.next = ListNode(v)
            curr = curr.next
        return head_new

11. 随机链表的复制

原题链接：138. 随机链表的复制 - 力扣（LeetCode）

class Node:
    def __init__(self, x, next=None, random=None):
        self.val = int(x)
        self.next = next
        self.random = random


class Solution(object):
    def copyRandomList(self, head):
        return copy.deepcopy(head)

12. 排序链表

原题链接：148. 排序链表 - 力扣（LeetCode）

class Solution(object):
    def sortList(self, head):
        if not head:
            return None
        lst = []
        while head:
            lst.append(head.val)
            head = head.next
        lst.sort()
         
        head_new = ListNode(int(lst[0]))
        curr = head_new
        for v in lst[1:]:
            curr.next = ListNode(int(v))
            curr = curr.next
        return head_new

13. 合并 K 个升序链表

原题链接：23. 合并 K 个升序链表 - 力扣（LeetCode）

class Solution(object):
    def mergeKLists(self, lists):
        if not lists:
            return None
        lst1 = []
        for head in lists:
            lst2 = []
            while head:
                lst2.append(head.val)
                head = head.next
            lst1.append(lst2)
        lst1 = sum(lst1, [])  
        if not lst1:
            return None      # lst1 = [[]]

        lst1.sort()
        head_new = ListNode(lst1[0])
        curr = head_new
        for v in lst1[1:]:
            curr.next = ListNode(v)
            curr = curr.next
        return head_new