刷题中的链表总结(持续跟新)

反转链表

注意：
链表问题伴随着大量指针操作。面试时不要急着写代码，仔细分析设计不容易出错。
由于链表是不连续存储的。当调节节点i的next指针时，需要（这里用tmp）来保存后节点的位置
常见错误：
① 输入链表头指针为null（特殊输入测试）或只有一个节点（一个功能测试）时，程序崩溃
②反转链表出现断裂
③返回的头节点不是原始链表的尾节点
解决方法：
提前想好测试用例，心中默默运行代码进行单元测试，再交给面试官

双指针循环解法**

# 
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution:
    # 返回ListNode
    def ReverseList(self, pHead):
        if not pHead or not pHead.next:# 判断特殊情况，链表为空或者只有一个node
            return pHead
        pre = None
        cur = pHead
        next = pHead.next
        while cur:# 注意这个地方是cur,还是cur.next，心中跑完代码再写
            tmp = cur.next
            cur.next = pre 
            pre = cur
            cur = tmp
        return pre

递归方法
递归方法我总是是是而非的，没有用一个时间段来用递归方法做题

class Solution:
    # 返回ListNode
    def ReverseList(self, pHead):
        if not pHead or not pHead.next:
            return pHead
        pNew = self.ReverseList(pHead.next)
        pHead.next.next = pHead
        pHead.next = None
        return pNew

两链表的第一个公共节点

思路：
hashmap可以解决不是面试官想要的思路，暴力时间复杂度O(MN)
双指针 第二种双指针好好看看！！！
创建两个指针p1和p2,分别指向两个链表的头结点，然后依次往后遍历。如果某个指针到达末尾，则将该指针指向另一个链表的头结点；如果两个指针所指的节点相同，则循环结束，返回当前指针指向的节点。比如两个链表分别为：1->3->4->5->6和2->7->8->9->5->6。短链表的指针p1会先到达尾部，然后重新指向长链表头部，当长链表的指针p2到达尾部时，重新指向短链表头部，此时p1在长链表中已经多走了k步(k为两个链表的长度差值)，p1和p2位于同一起跑线，往后遍历找到相同节点即可。其实该方法主要就是用链表循环的方式替代了长链表指针先走k步这一步骤

class Solution:# 傻逼双指针
    def FindFirstCommonNode(self, pHead1, pHead2):
        if not pHead1 or not pHead1.next or not pHead2 or not pHead2.next:
            return None
        length1 = 1
        length2 = 1
        tmp = 0
        cur1 = pHead1
        cur2 = pHead2
        while cur1:
            cur1 = cur1.next
            length1 += 1
        while cur2:
            cur2 = cur2.next
            length2 += 1
        if length1 > length2:
            while tmp < length1 - length2:
                pHead1 = pHead1.next
                tmp += 1
        if length1 <= length2:
            while tmp < length2 - length1:
                pHead2 = pHead2.next
                tmp += 1
        while pHead1:
            if pHead1.val == pHead2.val and pHead1.next == pHead2.next:
                return pHead1
            else:
                pHead1 = pHead1.next
                pHead2 = pHead2.next
        return None


```python#   巧妙精简化的双指针
# -*- coding:utf-8 -*-
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution:
    def FindFirstCommonNode(self, pHead1, pHead2):
        if not pHead1 or not pHead1.next or not pHead2 or not pHead2.next:
            return None
        p1 = pHead1
        p2 = pHead2
        while p1 != p2:
            p1 = p1.next if p1 else pHead2
            p2 = p2.next if p2 else pHead1
        return p1

链表中环的入口结点

hashmap

# -*- coding:utf-8 -*-
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution:
    def EntryNodeOfLoop(self, pHead):
        li = []# 刚开始我用的list命名：不能与关键词重合！
        p = pHead
        if not p or not p.next:
            return None
        while pHead:
            if pHead in li:# 列表查找O(N),所以这里换成dict 或者 set
            # if pHead in dict:
               # return pHead
           # else:
               # dict[pHead] = 1
           # pHead = pHead.next
                #return pHead
            else:
                li.append(pHead)# 是将pHead这个节点存到li中，而不是仅仅将pHead.val这个元素
            pHead = pHead.next
        return None

class Solution:# 用set
    def EntryNodeOfLoop(self, pHead):
        s = set([])
        while pHead:
            if pHead in s:
                return pHead
            else:
                s.add(pHead)
                pHead = pHead.next
        return None
        

快慢指针
思路：slow每次走一步，fast每次走两步
① 判断是否有环：如果slow追上fast(相遇)有环
② 找入口：让slow从相遇点出发，新指针从pHead出发，每次走一步，相遇即是入口（证明自己找找）

class Solution:
    def EntryNodeOfLoop(self, pHead):
        slow,fast = pHead,pHead
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
            if slow == fast:
                slow2 = pHead
                while slow != slow2:# 新指针从pHead出大
                    slow = slow.next
                    slow2 = slow2.next
                return slow
        return None

合并两个排序链表

双指针
注意！
tmp = ListNode(0)
return pHead.next

# -*- coding:utf-8 -*-
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution:
    # 返回合并后列表
    def Merge(self, pHead1, pHead2):
        tmp = ListNode(0)
        pHead = tmp
        while pHead1 and pHead2:# 注意
            if pHead1.val < pHead2.val:
                tmp.next = pHead1
                pHead1 = pHead1.next
            else:
                tmp.next = pHead2
                pHead2 = pHead2.next
            tmp = tmp.next
        if not pHead1:
            tmp.next = pHead2
        if not pHead2:
            tmp.next = pHead1
        return pHead.next
           

链表中倒数第K个节点

注意：
① 链表头节点为空指针
② 链表节点数小于k
③ k=0

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

class Solution:
    def kthToLast(self, head: ListNode, k: int) -> int:
        if not head or k < 0:
            return None
        a = head
        b = head
        for i in range(k):
            if not b:
                return None
            b = b.next
        while b:
            a = a.next
            b = b.next
        return a.val