Leetcode刷题笔记之专题1链表求解 Python实现

第一专题：链表求解

1. leetcode 224 反转链表(easy)

思路一：迭代

迭代需要三个指针，pre，cur，nxt，分别按顺序指向三个节点
三个指针的初始化：pre指向空节点，cur指向头结点head，nxt指向head.next
因为head.next可能不存在，nxt在循环中定义，这样如果head为空就不会进入循环
迭代过程

nxt指向cur.next
cur.next指向pre
pre移动到cur位置
cur移动到nxt位置

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

使用python多元赋值的特性求解：

先模拟出这个过程：使用1->2->3->4->5模拟作为例子

先初始化出：curr表示当前指针指向的对象为head；res代表结果指针,初始化为None。while循环进行res,curr赋值情况如下表：

class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        curr,res = head,None
        while curr:
            res,res.next,curr = curr,res,curr.next
        return res

次数	res = curr: res	res.next=res : res	curr = curr.next: curr
1	1->2->3->4->5	1->None	2->3->4->5
2	2->3->4->5	2->1->None	3->4->5
3	3->4->5	3->2->1->None	4->5
4	4->5	4->3->2->1->None	5
5	5	5->4->3->2->1->None	None

2. leetcode 21合并两个有序链表（easy)

一. 递归法：

# DeFinition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
        if l1 is None: return l2
        if l2 is None: return l1
        if l1.val < l2.val:
            l1.next = self.mergeTwoLists(l1.next,l2)
            return l1
        else:
            l2.next = self.mergeTwoLists(l1,l2.next)
            return l2

二. 迭代法

解决有序链表的一般解决思路，使用python的特性，先定义一个哨兵结点res，然后在遍历过程中维护一个pre指针，在循环迭代的过程中找到每一次的需求。

class Solution:
    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
        if l1 is None: return l2
        if l2 is None: return l1
        res = ListNode(0) # 首先预定义一个栈
        pre = ListNode(0)
        while l1 and l2:
            if l1.val > l2.val:
                pre.next = l2
                l2 = l2.next
            else:
                pre.next = l1
                l1 = l1.next
            pre = pre.next
        pre.next = l1 if l1 else l2
        return res.next

3. leetcode234 回文链表(easy)

一. 数组法

将链表的值复制到数组中后使用双指针法进行求解

class Solution:
    def ispalindrome(self, head: ListNode) -> bool:
        vals = []
        current_node = head
        while current_node is not None:
            vals.append(current_node.val)
            current_node = current_node.next
        return vals == vals[::-1]

时间复杂度O（n)，空间复杂度O（n), 其中链表转化为数组的时间复杂度为O(n)，空间复杂度也为O(n)。双指针数组对比时间复杂度为O(n)，空间复杂度也为O(n)。因此总的来看时间复杂度为O(n)，算法复杂度为O（n)。

二. 快慢指针法

。。。

4. leetcode26 删除有序数组中的重复项（easy)

使用双指针的方法：快指针用来判断前后是否相等，慢指针用来将不相等的值按顺序存储。

class Solution:
    def removeDuplicates(self, nums: List[int]) -> int:
        # 使用双指针的方法进行求解
        if not nums: return 0
        j = 0
        for i in range(1,len(nums)):
            if nums[i-1] != nums[i]:
                j +=1
                nums[j] = nums[i]
        return j+1

5. leetcode83. 删除排序链表中的重复元素

class Solution:
    def deleteDuplicates(self, head: ListNode) -> ListNode:
        curr = head
        if not head: return head
        while curr.next:
            if curr.val == curr.next.val:
                curr.next = curr.next.next
            else:
                curr = curr.next
        return head

6. leetcode61. 旋转链表

遍历出链表的长度
计算出链表需要从链表的哪个位置截断
将前半部分接到后半部分的尾部，然后进行输出

# DeFinition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def rotateRight(self, head: ListNode, k: int) -> ListNode:
        if not head or not head.next: return head
        curr = head
        n = 0
        while curr:
            n+=1
            curr = curr.next
        k = k % n
        if k == 0: return head
        p = head

        for i in range(n-k-1):
            p = p.next
        nxt = p.next
        p.next = None
        new_nxt = nxt

7. leetcode1669. 合并两个链表

# DeFinition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def mergeInBetween(self, list1: ListNode, a: int, b: int, list2: ListNode) -> ListNode:
        curr = list1
        for i in range(b+1):
            if i < a-1: curr = curr.next
            elif i == a-1:
                nxt = curr.next
                curr.next = list2
            else: nxt = nxt.next
        while curr.next:
            curr = curr.next
        curr.next = nxt
        return list1

8. leetcode23. 合并K个升序链表（hard)

假设链表的总长度为n，我们可以先得出k个升序链表的总长度，然后使用一个列表Docker存储当前K个链表的当前头节点的值，如果当前的链表为空，那么令其值为10**5（该值大于题目所规定的链表的最大值）。然后循环查找Docker中的最小值，并将其链接结果链表中。时间复杂度：O(KN), 空间复杂度O（K）

# DeFinition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def mergeKLists(self, lists: List[ListNode]) -> ListNode:
        res = ListNode(0)
        curr = res

        count = 0            # 查看所有链表的总长度count
        for item in lists:
            while item:
                count+= 1
                item = item.next
        if count == 0: return None
        docker = []
        for lst in lists:
                if lst is not None:
                    docker.append(lst.val)
                else:
                    docker.append(10**5) # 10**5代表无法取到的最大值

        while count > 0:     # 遍历count次将链表组合到res链表中
            count -= 1
            curr_min = min(docker)
            for i in range(len(docker)):
                if docker[i] == curr_min:
                    tmp = docker[i]
                    curr.next = lists[i]
                    curr = curr.next
                    if lists[i].next:
                        docker[i] = lists[i].next.val
                        lists[i] = lists[i].next
                    else:
                        docker[i] = 10**5
                    break
        return res.next