面试算法总结:链表
- 1、LeetCode----206. 链表翻转
-
- Solution1
- Solution2
- 2、LeetCode----24. 两两交换链表中的节点
-
- 基本思路
- 3、LeetCode----141. 环形链表
-
- 基本思路
- 4、LeetCode----142. 环形链表 II
-
- 基本思路
点击了解基本数据结构
1、LeetCode----206. 链表翻转
https://leetcode-cn.com/problems/reverse-linked-list/
Solution1
#基本思路
#首先将链表中的所有的数据读取存储到列表中
#然后结合ListNode,将列表的倒序填充到新的创建的链表中
#不断地插入后置节点,并不断的填充val,直至列表的所有数据填充完毕
class ListNode:
def __init__(self, x):
self.val = x
self.next = None
class Solution1:
def reverseList(self, head: ListNode) -> ListNode:
answer_val = []
if not head:
return head
while head:
answer_val.append(head.val)
head = head.next
answer_val = answer_val[::-1]
new_head = ListNode(answer_val[0])
answer_val.remove(answer_val[0])
p = new_head
for i in answer_val:
temp = ListNode(i)
p.next = temp
p = p.next
return new_head
Solution2
#基本思路
#直接利用链表的特性进行构造将链表的前驱和后继互换即可
class Solution2:
def reverseList(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
p, rev = head, None
while p:
rev, rev.next, p = p, rev, p.next
return rev
2、LeetCode----24. 两两交换链表中的节点
https://leetcode-cn.com/problems/swap-nodes-in-pairs/submissions/
基本思路
#利用一个flag值将链表区分为奇数和偶数
#对链表进行分组,每两个进行一组
#不断地交换每一组中的前后两个数据
#奇数时最后一个元素不进行交换
#Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def swapPairs(self, head: ListNode) -> ListNode:
p, v = head, None
flag = 0
if not head or not head.next:
return head
while p:
try:
if not flag % 2:
v = p.next
p.val, v.val = v.val, p.val
except:
pass
flag += 1
p = p.next
return head
3、LeetCode----141. 环形链表
https://leetcode-cn.com/problems/linked-list-cycle/
基本思路
#利用两个指针,分别指向链表的头部,两个指针以不同的速度向前移动
#一个移动一次移动一位,一个一次移动两位
#最后判断,如果存在两个指针的后继指向同一个节点,那么说明此为循环链表。
#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
"""
other_head = head
while head:
try:
head = head.next
other_head = other_head.next.next
if head.next == other_head.next:
return True
except:
return False
return False
4、LeetCode----142. 环形链表 II
https://leetcode-cn.com/problems/linked-list-cycle-ii/
基本思路
#利用set数据结构的唯一性,将已经访问过的数据存入的set表中
#当出现head在set表中时此时说明此节点为再次出现的节点
#所以这个节点就肯定是我们要找的环形链表的起点
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def detectCycle(self, head):
set1=set([])
while head not in set1:
set1.add(head)
if not head:
break
head=head.next
return head