Reverse a singly linked list. Example: Input: 1->2->3->4->5->NULL Output: 5->4->3->2->1->NULL # Definition for singly-linked list. # class ListNode(object): # def __init__(self, x): # self.val = x # self.next = None class Sol…

# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution(object):    def reverseList(self, head):        if(head==None):return None        if(head.next==No…

Given a string, you need to reverse the order of characters in each word within a sentence while still preserving whitespace and initial word order. Example 1: Input: "Let's take LeetCode contest" Output: "s'teL ekat edoCteeL tsetnoc"…

Given a string and an integer k, you need to reverse the first k characters for every 2k characters counting from the start of the string. If there are less than k characters left, reverse all of them. If there are less than 2k but greater than or eq…

Given a string S, return the "reversed" string where all characters that are not a letter stay in the same place, and all letters reverse their positions.   Example 1: Input: "ab-cd" Output: "dc-ba" Example 2: Input: "a-…

206. Reverse Linked List 之前在牛客上的写法: 错误代码: class Solution { public: ListNode* ReverseList(ListNode* pHead) { if(pHead == NULL) return NULL; ListNode* p1 = pHead; ListNode* p2 = pHead->next; ListNode* p3 = pHead->next->next; pHead->next = NULL;…

Question 206. Reverse Linked List Solution 题目大意:对一个链表进行反转 思路: Java实现: public ListNode reverseList(ListNode head) { ListNode newHead = null; while (head != null) { ListNode tmp = head.next; head.next = newHead; newHead = head; head = tmp; } return new…

表不支持随机查找,通常是使用next指针进行操作. 206. 反转链表 /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ //时间:O(n) 只遍历了一遍链表 //空间:O(1) 开了三个指针的空间 class Solution { public: ListNode*…

206. Reverse Linked List[easy] Reverse a singly linked list. Hint: A linked list can be reversed either iteratively or recursively. Could you implement both? 解法一: class Solution { public: ListNode* reverseList(ListNode* head) { ListNode * pre = NULL;…

Reverse a singly linked list. Hint: A linked list can be reversed either iteratively or recursively. Could you implement both? 反向链表,分别用递归和迭代方式实现. 递归Iteration: 新建一个node(value=任意值, next = None), 用一个变量 next 记录head.next,head.next指向新node.next,新 node.next…