【经典LeetCode算法题目专栏分类】【第7期】快慢指针与链表

《博主简介》

小伙伴们好,我是阿旭。专注于人工智能AI、python、计算机视觉相关分享研究。

更多学习资源,可关注公-仲-hao:【阿旭算法与机器学习】,共同学习交流~

👍感谢小伙伴 们点赞、关注!

快慢指针

移动零

|-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|
| class Solution : def moveZeroes****(**** self****,**** nums****:**** List****[**** int ]) -> None : """ Do not return anything, modify nums in-place instead. """ left = 0 n = len ( nums****)**** for i in range ( n****):**** if nums****[**** i****]**** != 0****:**** nums****[**** left****],**** nums****[**** i****]**** = nums****[**** i****],**** nums****[**** left****]**** left += 1 return nums class Solution : def moveZeroes****(**** self****,**** nums****:**** List****[**** int ]) -> None : """ Do not return anything, modify nums in-place instead. """ j = 0 for i in range ( len ( nums****)):**** if nums****[**** i****]**** != 0****:**** nums****[**** j****]**** = nums****[**** i****]**** if i != j****:**** nums****[**** i****]**** = 0 j += 1 return nums |

链表

两两交换链表中的节点

|-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|
| # 迭代 class Solution : def swapPairs****(**** self****,**** head****:**** ListNode****)**** -> ListNode****:**** # 通过迭代实现 dummy = ListNode****(-**** 1****)**** dummy****.**** next = head prev_node = dummy while head and head****.**** next : first_node = head second_node = head****.**** next # 交换节点 prev_node****.**** next = second_node first_node****.**** next = second_node****.**** next second_node****.**** next = first_node # 初始化头节点与prev_node prev_node = first_node head = first_node****.**** next return dummy****.**** next # 递归 class Solution : def swapPairs****(**** self****,**** head****:**** ListNode****)**** -> ListNode****:**** # 递归实现 if not head or not head****.**** next : return head first_node = head second_node = head****.**** next # 第二个节点的next节点作为头部传入递归函数,返回的是 # 指向第二个节点的指针 first_node****.**** next = self****.**** swapPairs****(**** second_node****.**** next ) second_node****.**** next = first_node return second_node |

反转链表

将链表进行反转

|-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|
| # 迭代 class Solution : def reverseList**(** self**,** head**:** ListNode**)** -> ListNode**:** if head is None : return head pre = None cur = head while cur**:** nxt = cur**.** next cur**.** next = pre pre = cur cur = nxt return pre # 递归 class Solution : def reverseList**(** self**,** head**:** ListNode**)** -> ListNode**:** if not head or not head**.** next : return head last = self**.** reverseList**(** head**.** next ) head**.** next . next = head head**.** next = None return last |

关于本篇文章大家有任何建议或意见,欢迎在评论区留言交流!

觉得不错的小伙伴,感谢点赞、关注加收藏哦!
欢迎关注下方GZH:阿旭算法与机器学习,共同学习交流~

相关推荐
爱学java的ptt3 小时前
206反转链表
数据结构·链表
FPGA_无线通信3 小时前
OFDM 同步设计(3)
算法·fpga开发
SHOJYS3 小时前
离散化+二位前缀和的计数题 [USACO20DEC] Rectangular Pasture S
算法
java修仙传3 小时前
力扣hot100:最大子数组和
数据结构·算法·leetcode
想唱rap3 小时前
C++之unordered_set和unordered_map
c++·算法·哈希算法
Rock_yzh3 小时前
LeetCode算法刷题——54. 螺旋矩阵
数据结构·c++·学习·算法·leetcode·职场和发展·矩阵
papership4 小时前
【入门级-算法-5、数值处理算法:高精度整数除以单精度整数的商和余数】
算法
CoderYanger4 小时前
C.滑动窗口-求子数组个数-越短越合法——3258. 统计满足 K 约束的子字符串数量 I
java·开发语言·算法·leetcode·1024程序员节
2301_807997384 小时前
代码随想录-day56
算法
AI科技星4 小时前
时空运动的几何约束:张祥前统一场论中圆柱螺旋运动光速不变性的严格数学证明与物理诠释
服务器·数据结构·人工智能·python·科技·算法·生活