Leetcode之链表（前200道）

Posted 2020-10-13

Easy

1. Merge Two Sorted Lists:

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

这道题有两种解法，一种是iterator，一种是recursively

iterator:

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

class Solution(object):
    def mergeTwoLists(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        result=current=ListNode(0)
        if not l1:
            return l2
        if not l2:
            return l1
        while l1 and l2:
            if l1.val<l2.val:
                current.next=l1
                l1=l1.next
            else:
                current.next=l2
                l2=l2.next
            current=current.next
        current.next=l1 or l2
        return result.next

recursively:

class Solution(object):
    def mergeTwoLists(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        if not l1: 
            return l2
        if not l2:
            return l1
        if l1.val==l2.val:
            l1.next=self.mergeTwoLists(l1.next,l2)
            return l1
        if l1.val<l2.val:
            l1.next=self.mergeTwoLists(l1.next,l2)
            return l1
        if l1.val>l2.val:
            l2.next=self.mergeTwoLists(l2.next,l1)
            return l2

这道题用了两种方法：recursively和iterator。 recursively比较难理解，画出stack模型会比较容易理解，主要是通过反复调用方法来实现。 iterator比较容易理解，主要思想是创造两个新的链表，result和current，current负责存储通过比较l1和l2得出的结果的节点， result负责最后的返回，result=current=ListNode(0)。要注意：每次比较完之后，要用l1=l1.next来移动到下一个节点进行比较 同时，每一次循环最后都需要current=current.next来移动到下一个节点。当l1或者l2的节点为none时，跳出循环，使用current.next=l1orl2 来将另外一个节点剩余节点存入current，最后return result.next

2. Remove Duplicates from Sorted List:

Given a sorted linked list, delete all duplicates such that each element appear only once.

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

class Solution(object):
    def deleteDuplicates(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        current=head
        if not current:
            return None
        while current.next:
            if current.next.val==current.val:
                current.next=current.next.next
            else:
                current=current.next
        return head

1.single-linked list的head不是空
2.最后返回是return head，可以返回整个list

3. Linked List Cycle:
Given a linked list, determine if it has a cycle in it.

# 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
        """
        try:
            slow=head
            fast=head.next
            while slow is not fast:
                slow=slow.next
                fast=fast.next.next
            return True
        except:
            return False
            
            
        

这道题要mark一下
很有意思的算法，设了两个指针，一个快指针，一个慢指针，快指针每次.next.next，慢指针.next，用了try...except，如果是有cycle的话，
慢指针和快指针总会遇到，于是返回true
如果没有（except），返回false

4.Intersection of Two Linked Lists:
Write a program to find the node at which the intersection of two singly linked lists begins.

# 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
        """
        a=headA
        b=headB
        if a is None or b is None:
            return None

        while a is not b:
            if a is None:
                a=headB
            else:
                a=a.next
            if b is None:
                b=headA
            else:
                b=b.next
        return a
    

这个算法也要mark一下！很机智！
核心就是指向两个链表节点的指针一起移动，但是最tricky的地方在于，由于两个链表长度不同，到达intersection的时间也不同，所以可能跑完整个链表
都不会相遇，所以每个链表跑完之后（=none），则指向另一个链表的head，继续跑，这样每次都在缩短两个链表到达intersection的距离，于是最后在
intersection完美相遇！