ListNode *rotateRight(ListNode *head, int k) { if (!head ) return head; if (k == 0) return head; vector<ListNode *> nodeVec; for (ListNode *curr = head; curr != NULL; curr = curr->next) nodeVec.push_back(curr); int listLen = nodeVec.size(); k %= listLen; if (k ==...
1、旋转链表:Rotate List - LeetCodeGiven a linked list, rotate the list to the right by kplaces, wherekis non-negative.class Solution { public: ListNode* rotateRight(ListNode* … Uno Whoiam 刷了LeetCode的链表专题,我发现了一个秘密! Drink凉白开 LeetCode 题解 | 19. 删除链表的倒数第N个节点...
Given a list, rotate the list to the right by k places, where k is non-negative. For example: Given 1->2->3->4->5->NULL and k = 2, return 4->5->1->2->3->NULL. 这题要求把链表后面k个节点轮转到链表前面。 对于这题我们首先需要遍历链表,得到链表长度n,因为k可能大于n,所以我们...
代码(Python3) # Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: def rotateRight(self, head: Optional[ListNode], k: int) -> Optional[ListNode]: # 如果链表为空或者 k 为 0 ,直接...
这道题的rotate的定义很有趣,是把最后一个元素挪到了最前面去,然后这样执行K次。 我们可以先遍历一遍链表,计算得到链表的长度n,然后把首尾两个节点相连。 从尾节点走(n - k%n) 得到的节点的下一个节点就是新的节点了! 代码: /** * Definition for singly-linked list. ...
# Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: def rotateRight(self, head, k): """ :type head: ListNode :type k: int :rtype: ListNode """ # 边界情况处理 if not head: return head # 统计...
* Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode *rotateRight(ListNode *head, int k) { if(head == NULL || k == 0) return head; ListNode* cur = he...
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 */ 9 class Solution { 10 public: 11 ListNode* rotateRight(ListNode* head, int k) { 12 if(head == NULL) ...
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode* rotateRight(ListNode* head, int k) { if(head==NULL) ...
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode *rotateRight(ListNode *head, int k) { // Start typing your C/C++ solution below // DO NOT write...