十五天的时间,刷完了所有的简单题,避免遗忘,所以开始简单题的二刷,第一遍刷题的时候过得速度比较快,因为我觉得基础不好的我,不要硬着头皮去想最优的方法,而是应该尽量去学一些算法思想,所以每道题只给自己5-10分钟的时间想,想不出来的就去找相关的答案,所以刷的比较快。二刷的时候按照leetcode官方给出的题目分类展开,同时,将解题思路记录于加深印象。
想要一起刷题的小伙伴,我们一起加油吧!
我的github连接:https://github.com/princewen/leetcode_python
21. Merge Two Sorted Lists
很简单的链表拼接题,但是要注意两个地方
1、返回值要返回head.next
2、无需判断循环后哪个不为空,or返回第一个为真的值
# 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
"""
head = cur = ListNode(0)
while l1 and l2:
if l1.val > l2.val:
cur.next = l2
l2 = l2.next
else:
cur.next = l1
l1 = l1.next
cur = cur.next
cur.next = l1 or l2
return head.next
83. Remove Duplicates from Sorted List
从链表中删除元素,一般需要两个指针来完成,分别记录要删除的链表元素和上一个元素。按照这个思路:
# 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
"""
if not head:
return head
p = head
q = head.next
while q:
if q.val == p.val:
p.next = q.next
q = q.next
else:
p = p.next
q = q.next
return head
两个指针比较容易理解,但是这里其实用一个指针就可以完成:
# 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
"""
cur = head
while cur:
while cur.next and cur.val == cur.next.val:
cur.next = cur.next.next
cur = cur.next
return head
141. Linked List Cycle
判断链表中是否是有环,采用追赶法的思路,设置一个walker,每次走一步,设置一个runner,每次跑两步,当runner追上walkder时,说明链表中有环存在。
# 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
"""
if not head:
return False
walker = head
runner = head.next
try:
while walker!=runner:
walker = walker.next
runner = runner.next.next
return True
except:
return False
160. Intersection of Two Linked Lists
判断链表是否有交集,可以设置两个指针,一个指针从第一个链表开始遍历,遍历完第一个链表再遍历第二个链表,另一个指针从第二个链表开始遍历,遍历完第二个链表再遍历第一个链表,不管两个链表在交集前的长度如何,如果有交集的话,两个指针肯定会同时遍历到最后的交集部分。
# 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
"""
if not headA or not headB:
return None
pa = headA
pb = headB
while pa is not pb:
pa = headB if pa == None else pa.next
pb = headA if pb == None else pb.next
return pa
203. Remove Linked List Elements
"有了第83题的思路,我们这里可以用一个指针来进行链表的遍历,但是这里需要注意的是,头节点也需要进行判断,如果头节点的值等于val的话,我们不能返回头节点,所以这里很巧妙的重新生成了一个无关的头节点。
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def removeElements(self, head, val):
"""
:type head: ListNode
:type val: int
:rtype: ListNode
"""
dummy = ListNode(-1)
dummy.next = head
cur = dummy
while cur:
while cur.next and cur.next.val == val:
cur.next = cur.next.next
cur=cur.next
return dummy.next
206. Reverse Linked List
链表转置,这里实在想不到一个指针的解法了,只能用两个指针,再加上head的帮忙,p指针记录的是每次的队头元素,q指针指向下一个要插入队头的元素。
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def reverseList(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
if not head:
return None
p = head
q = head.next
while q:
head.next = q.next
q.next = p
p = q
q = head.next
return p
234. Palindrome Linked List
判断一个链表是否是回文的,很自然的想法就是两个指针,一个指针从前往后走,一个指针从后往前走,判断元素值是否相同,这里要分几个步骤来进行求解:
1、找到链表长度的一半,用追赶法,一个指针一次走两步,一个指针一次走一步
2、将后一半数组转置
3、判断链表是否是回文链表
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def isPalindrome(self, head):
"""
:type head: ListNode
:rtype: bool
"""
slow = fast = head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
node = None
while slow:
nxt = slow.next
slow.next = node
node = slow
slow = nxt
while node and head:
if node.val != head.val:
return False
node = node.next
head = head.next
return True
237. Delete Node in a Linked List
这道题关键是理解题意,不给你整个链表,只给你一个节点,如何把这个节点删除,其实我们没必要真的把这个节点删除,而是把这个节点对应的val值删除即可,所以我们可以偷天换日,把下一个节点的值赋给这个节点,再把下一个节点删除。
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def deleteNode(self, node):
"""
:type node: ListNode
:rtype: void Do not return anything, modify node in-place instead.
"""
node.val = node.next.val
node.next = node.next.next
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