摘要:而后吾当依除取余之法,化大为小,则指针不致于越界也。后欲寻右起第结点,令快指针先行数日,及至两指针相距为,便吟鞭东指,与慢指针策马共进。快慢指针亦止于其所焉。舞动长剑,中宫直入,直取首级,而一掌劈空,已鸿飞冥冥。自此,一代天骄,霸业已成。
Problem
Given a list, rotate the list to the right by k places, where k is non-negative.
ExampleGiven 1->2->3->4->5 and k = 2, return 4->5->1->2->3.
Note这是一道有故事的题目。
k有多大?其翼若垂天之云,则或长于链表。链表几长?北海也,天池也,其广数千里,则k亦可容于链表也。
故吾必先穷尽链表之长度,如若head为null,抑或其长可为k所整除,则此题不必再做,返回head可也。
而后吾当依除len取余之法,化大k为小k,则指针不致于越界也。后欲寻右起第k结点,令快指针先行数日,及至两指针相距为k,便吟鞭东指,与慢指针策马共进。则快指针行至null时,慢指针可至右起第k处矣。
方是时也,孤灯长夜,指飞键落。如风吹败叶,雨打窗棂。快慢指针亦止于其所焉。此时看官不妨温酒一盏,看那链表翻转,指针易位。那新头目唤作curhead,正是slow所指之人。fast舞动长剑,中宫直入,直取head首级,而slow一掌劈空,null已鸿飞冥冥。
自此,curhead一代天骄,霸业已成。
public class Solution { public ListNode rotateRight(ListNode head, int k) { ListNode fast = head, slow = head; int len = 0; while (fast != null) { fast = fast.next; len++; } if (head == null || k % len == 0) return head; fast = head; k = k % len; while (k-- > 0) fast = fast.next; while (fast.next != null) { fast = fast.next; slow = slow.next; } ListNode curhead = slow.next; slow.next = null; fast.next = head; return curhead; } }
文章版权归作者所有,未经允许请勿转载,若此文章存在违规行为,您可以联系管理员删除。
转载请注明本文地址:https://www.ucloud.cn/yun/65689.html
Problem Given an array, rotate the array to the right by k steps, where k is non-negative. Example Example 1: Input: [1,2,3,4,5,6,7] and k = 3Output: [5,6,7,1,2,3,4]Explanation:rotate 1 steps to the r...
摘要:两种方法,转置镜像法和公式法。首先看转置镜像法原矩阵为转置后水平镜像翻转后所以,基本的思路是两次遍历,第一次转置,第二次水平镜像翻转变换列坐标。公式法是应用了一个翻转的公式如此翻转四次即可。二者均可,并无分别。 Problem You are given an n x n 2D matrix representing an image.Rotate the image by 90 de...
Problem Given an array of meeting time intervals consisting of start and end times [[s1,e1],[s2,e2],...] (si < ei), determine if a person could attend all meetings. Example Given intervals = [[0,30],[...
摘要:依然是一道找倒数第个结点的链表题,用双指针做。先走,然后和一起走,直到为,的位置就是倒数第个位置。 Problem Find the nth to last element of a singly linked list. The minimum number of nodes in list is n. Example Given a List 3->2->1->5->null ...
摘要:大体意思就是,先复制到,顺便将所有的放在再复制所有的到,顺便将所有的放在最后令,令,将和分离,返回的头结点 Problem A linked list is given such that each node contains an additional random pointer which could point to any node in the list or null. ...
阅读 2064·2021-11-24 09:39
阅读 1496·2021-10-11 10:59
阅读 2455·2021-09-24 10:28
阅读 3340·2021-09-08 09:45
阅读 1242·2021-09-07 10:06
阅读 1579·2019-08-30 15:53
阅读 2032·2019-08-30 15:53
阅读 1395·2019-08-30 15:53