leetcode-reverse linked list

本文详细解析了如何使用C++及Python实现链表的反转算法。通过实例输入1->2->3->4->5->NULL,展示其反转过程,输出为5->4->3->2->1->NULL。代码中提供了迭代方法,包括初始化当前节点、下一个节点和前一个节点,通过循环操作完成链表反转。

题目 reverse linked list


描述

Reverse a singly linked list.
Example:
Input: 1->2->3->4->5->NULL
Output: 5->4->3->2->1->NULL

解法
c++ 慢慢

class Solution {
public:
ListNode* reverseList(ListNode* head) {
ListNode* curr = head;
ListNode* next = NULL;
ListNode* prev = NULL;
while (curr != NULL) {
next = curr->next;
curr->next = prev;
prev = curr;
curr = next;
}
return prev;
}
};


python

class Solution:
def reverseList(self, head):
“”"
:type head: ListNode
:rtype: ListNode
“”"
curr = head
prev = None
next = None
while curr:
next = curr.next
curr.next=prev
prev = curr
curr = next
return prev

以下是几种 LeetCode 234 题回文链表问题的 Python 实现: ### 方法一:将链表复制到数组里再从两头比对 ```python # Definition for singly-linked list. class ListNode: def __init__(self, val=0, next=None): self.val = val self.next = next class Solution: def isPalindrome(self, head: ListNode) -> bool: lst = [] node = head while node: lst.append(node.val) node = node.next start = 0 end = len(lst) - 1 while start < end: if lst[start] != lst[end]: return False start += 1 end -= 1 return True ``` ### 方法二:递归法 ```python # Definition for singly-linked list. class ListNode(object): def __init__(self, val=0, next=None): self.val = val self.next = next class Solution(object): def isPalindrome(self, head): front_pointer = head def recursively_check(current_node=head): if current_node is not None: if not recursively_check(current_node.next): return False if front_pointer.val != current_node.val: return False nonlocal front_pointer front_pointer = front_pointer.next return True return recursively_check() ``` ### 方法三:快慢指针 + 反转链表 ```python # Definition for singly-linked list. class ListNode: def __init__(self, x): self.val = x self.next = None class Solution: def isPalindrome(self, head: ListNode) -> bool: if head is None: return True first_half_end = self.end_of_first_half(head) second_half_start = self.reverse_list(first_half_end.next) result = True first_position = head second_position = second_half_start while result and second_position is not None: if first_position.val != second_position.val: result = False first_position = first_position.next second_position = second_position.next first_half_end.next = self.reverse_list(second_half_start) return result def end_of_first_half(self, head): fast = head slow = head while fast.next is not None and fast.next.next is not None: fast = fast.next.next slow = slow.next return slow def reverse_list(self, head): previous = None current = head while current is not None: next_node = current.next current.next = previous previous = current current = next_node return previous ```
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