Leetcode之鏈表(前200道)
阿新 • • 發佈:2017-11-09
ica div eth color exc type rmi 鏈表 res
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.
這道題有兩種解法,一種是iterator,一種是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 ifnot 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 returnresult.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
這道題用了兩種方法:recursively和iterator。 recursively比較難理解,畫出stack模型會比較容易理解,主要是通過反復調用方法來實現。 iterator比較容易理解,主要思想是創造兩個新的鏈表,result和current,current負責存儲通過比較l1和l2得出的結果的節點, result負責最後的返回,result=current=ListNode(0)。要註意:每次比較完之後,要用l1=l1.next來移動到下一個節點進行比較 同時,每一次循環最後都需要current=current.next來移動到下一個節點。當l1或者l2的節點為none時,跳出循環,使用current.next=l1orl2 來將另外一個節點剩余節點存入current,最後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.最後返回是return head,可以返回整個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
這道題要mark一下
很有意思的算法,設了兩個指針,一個快指針,一個慢指針,快指針每次.next.next,慢指針.next,用了try...except,如果是有cycle的話,
慢指針和快指針總會遇到,於是返回true
如果沒有(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
這個算法也要mark一下!很機智!
核心就是指向兩個鏈表節點的指針一起移動,但是最tricky的地方在於,由於兩個鏈表長度不同,到達intersection的時間也不同,所以可能跑完整個鏈表
都不會相遇,所以每個鏈表跑完之後(=none),則指向另一個鏈表的head,繼續跑,這樣每次都在縮短兩個鏈表到達intersection的距離,於是最後在
intersection完美相遇!
Leetcode之鏈表(前200道)