publicclassSolution {publicList<String>restoreIpAddresses(String s) { List<String> res =newArrayList<String>();intlen =s.length();for(inti = 1; i<4 && i<len-2; i++){for(intj = i+1; j<i+4 && j<len-1; j++){for(intk = j+1; k<j+4 && k<len; k++){ String s1= s....
Reverse a singly linked list. 做II之前应该先来做1的,这个倒是很简单,基本上不用考虑什么,简单的链表反转而已: 1/**2* Definition for singly-linked list.3* struct ListNode {4* int val;5* ListNode *next;6* ListNode(int x) : val(x), next(NULL) {}7* };8*/9classSolution {10public:11...
(参考视频讲解:Leetcode力扣|206反转链表|递归|reverse linked list_哔哩哔哩_bilibili) # 定义一个链表节点类classListNode:def__init__(self,val=0,next=None):# 初始化函数self.val=val# 节点的值self.next=next# 指向下一个节点的指针# 将给出的数组转换为链表deflinkedlist(list):head=ListNode(list[0]...
https://leetcode-cn.com/problems/reverse-linked-list/solution/shi-pin-jiang-jie-die-dai-he-di-gui-hen-hswxy/ python # 反转单链表 迭代或递归 class ListNode: def __init__(self, val): self.val = val self.head = None class Solution: def reverseList2(self, head: ListNode) -> ListNode...
[LeetCode]Reverse Linked List Question Reverse a singly linked list. 本题难度Easy。 3指针法 复杂度 时间O(N) 空间 O(1) 思路 利用3个指针对链表实施reverse。 代码 /** * Definition for singly-linked list. * public class ListNode { * int val;...
Leetcode260反转链表(java/c++/python) JAVA: class Solution { public ListNode reverseList(ListNode head) { if( head == null || head.next == null) return head; ListNode newHead = reverseList(head.next); head.next.next = head; head.next = null; return newHead; } } C++: class Solution...
LeetCode 206. 反转链表(Reverse Linked List) 示例: 输入:1->2->3->4->5->NULL输出:5->4->3->2->1->NULL 切题 一、Clarification 只需注意为空链表的情况 二、Possible Solution 1、迭代 2、递归 可利用哨兵简化实现难度 Python3 实现
https://leetcode.com/problems/reverse-linked-list/discuss/140916/Python-Iterative-and-Recursive recursive method 每次把head.next保存下来,这样就不用像我之前写的那样,用while循环去找最后一个node了,大大降低了时间复杂度 直接将linked list的值存入stack就好,不需要把整个node存进去。只存值的好处是后面可以...
publicclassSolution{publicListNodereverseBetween(ListNode head,int m,int n){if(m==n||head==null||head.next==null){returnhead;}ListNode pre=newListNode(0);pre.next=head;ListNode Mpre=pre;ListNode NodeM=head;ListNode NodeN=head;int mNum=1;int nNum=1;while(mNum<m&&NodeM!=null){Mpre=...
Can you solve this real interview question? Reverse Linked List - Given the head of a singly linked list, reverse the list, and return the reversed list. Example 1: [https://assets.leetcode.com/uploads/2021/02/19/rev1ex1.jpg] Input: head = [1,2,3,