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...
nodeVec.push_back(curr); int listLen = nodeVec.size(); k %= listLen; if (k == 0) return head; nodeVec[listLen - 1]->next = nodeVec[0]; nodeVec[listLen - k - 1]->next = NULL; return nodeVec[listLen - k]; } 不管是左旋还是右旋,均可通过上面的方式来实现。 空间复杂度变...
leetcode -- Rotate List -- 重点 这里思路很简单。但是要注意,k可以大于len(linkedlist). 要求得长度之后取mod。还有就是这个rotate,不需要把rotate部分逆置。 def rotateRight(self, head, k): """ :type head: ListNode :type k: int :rtype: ListNode """ if not head or not head.next: return ...
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. * struct ListNode { * int val;...
【leetcode】61. Rotate List 题目说明 https://leetcode-cn.com/problems/rotate-list/description/ 给定一个链表,旋转链表,将链表每个节点向右移动 k 个位置,其中 k 是非负数。 解法1 解法2...Leetcode 61. Rotate List Given a linked list, rotate the list to the right by k places, where k ...
Leetcode 61 Rotate List Leetcode 61 Rotate List 题目 思路 代码 题目 Given a linked list, rotate the list to the right by k places, where k is non-negative. Example 1: Input: 1->2->3->4->5->NULL, k = 2 Output: 4->5-......
LeetCode 0061 - Rotate List list Given a list, rotate the list to the right by k places, where k is non-negative. Reck Zhang 2021/08/11 3040 LinkedList - 61. Rotate List 编程算法 Given a linked list, rotate the list to the right by k places, where k is non-negative. ppxai 2020...
参考:Rotate List leetcode java /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: /**
[LeetCode]--61. Rotate List 简介: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 Given a list, rotate the list to the right by k places, where k is non-...
packageleetcodefuncsearch33(nums[]int,targetint)int{iflen(nums)==0{return-1}low,high:=0,len(nums)-1forlow<=high{mid:=low+(high-low)>>1ifnums[mid]==target{returnmid}elseifnums[mid]>nums[low]{// 在数值大的一部分区间里ifnums[low]<=target&&target<nums[mid]{high=mid-1}else{low...