【经典LeetCode算法题目专栏分类】【第7期】快慢指针与链表

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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 |

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