LeetCode 206-Reverse Linked List

本文介绍了一种反转单链表的方法,特别关注了处理特殊情况的细节,例如链表长度为0、1或2的情况。通过一个具体的Java代码示例,详细展示了如何通过迭代的方式实现链表的反转。

题目:Reverse a singly linked list.

分析:此题在于细节,处理链表长度为0、1、2长度的特殊情况

代码:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode reverseList(ListNode head) {
        if(head==null){
            return null;
        }
        ListNode left ,middle,right;
        left=head;
        middle=left.next;
        if(middle==null){
        	return left;
        }
        right=middle.next;
        left.next=null;
        while(right!=null){
    	  
    	   middle.next=left;
    	   left=middle;
    	   middle=right;
    	   right=middle.next;
    	   
       }
        middle.next=left;
        
		return middle;
        
        
    }
}

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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