在线时间:8:00-16:00
迪恩网络APP
随时随地掌握行业动态
扫描二维码
关注迪恩网络微信公众号
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★ 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->1->2->3->NULL Explanation: rotate 1 steps to the right: 5->1->2->3->4->NULL rotate 2 steps to the right: 4->5->1->2->3->NULL Example 2: Input: 0->1->2->NULL, k = 4 Output: 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。 示例 1: 输入: 1->2->3->4->5->NULL, k = 2 输出: 4->5->1->2->3->NULL 解释: 向右旋转 1 步: 5->1->2->3->4->NULL 向右旋转 2 步: 4->5->1->2->3->NULL 示例 2: 输入: 0->1->2->NULL, k = 4 输出: 20ms 1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * public var val: Int 5 * public var next: ListNode? 6 * public init(_ val: Int) { 7 * self.val = val 8 * self.next = nil 9 * } 10 * } 11 */ 12 class Solution { 13 func rotateRight(_ head: ListNode?, _ k: Int) -> ListNode? { 14 // Count length 15 var l = 0 16 var p = head 17 var last = p 18 while p != nil { 19 l += 1 20 last = p 21 p = p!.next 22 } 23 24 if l == 0 { return head } 25 26 // Rotate 27 let dummyHead = ListNode(0) 28 dummyHead.next = head 29 p = dummyHead 30 for _ in 0..<(l - k % l) % l { 31 p = p!.next 32 } 33 let ret = p!.next 34 p!.next = nil 35 last!.next = dummyHead.next 36 return ret 37 } 38 } 24ms 1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * public var val: Int 5 * public var next: ListNode? 6 * public init(_ val: Int) { 7 * self.val = val 8 * self.next = nil 9 * } 10 * } 11 */ 12 class Solution { 13 func rotateRight(_ head: ListNode?, _ k: Int) -> ListNode? { 14 if(head?.next == nil || k == 0){ 15 return head 16 } 17 var current = head 18 var array = [Int]() 19 var n = 0 20 while(current != nil){ 21 array.append(current!.val) 22 n += 1 23 current = current!.next 24 } 25 var preHead = ListNode(0) 26 current = preHead 27 var start = (n - (k % n)) % n 28 for i in start ..< (start + n){ 29 var newNode = ListNode(array[(i % n)]) 30 current!.next = newNode 31 current = current!.next 32 } 33 return preHead.next 34 } 35 } 28ms 1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * public var val: Int 5 * public var next: ListNode? 6 * public init(_ val: Int) { 7 * self.val = val 8 * self.next = nil 9 * } 10 * } 11 */ 12 class Solution { 13 func rotateRight(_ head: ListNode?, _ k: Int) -> ListNode? { 14 if head == nil || head?.next == nil || k == 0 { 15 return head 16 } 17 var (fastPtr, slowPtr, length) = (head, head, 1) 18 while let _ = fastPtr?.next { // get list length 19 length += 1 20 fastPtr = fastPtr?.next 21 } 22 23 let slowLenth = length - k%length 24 for _ in 1..<slowLenth { // how many steps slowPtr needs to go 25 slowPtr = slowPtr?.next 26 } 27 28 fastPtr?.next = head // perform rotation 29 let newHead = slowPtr?.next 30 slowPtr?.next = nil 31 32 return newHead; 33 } 34 } 28ms 1 class Solution { 2 func rotateRight(_ head: ListNode?, _ k: Int) -> ListNode? { 3 guard let head = head else { 4 return nil 5 } 6 var newHeadNode: ListNode! = head; 7 var newTailNode: ListNode! = head; 8 var tailNode: ListNode! = head; 9 var linkLength = 1; 10 while tailNode.next != nil { 11 tailNode = tailNode.next; 12 linkLength += 1 13 } 14 if linkLength <= 1 { 15 return head; 16 } 17 let k = k % linkLength 18 for _ in 0 ..< abs(linkLength - k - 1) { 19 if newTailNode.next != nil { 20 newTailNode = newTailNode.next 21 } else { 22 newTailNode = head 23 } 24 } 25 if newTailNode.next != nil { 26 newHeadNode = newTailNode.next 27 tailNode.next = head; 28 newTailNode.next = nil; 29 return newHeadNode; 30 } else { 31 return head; 32 } 33 } 34 } 40ms 1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * public var val: Int 5 * public var next: ListNode? 6 * public init(_ val: Int) { 7 * self.val = val 8 * self.next = nil 9 * } 10 * } 11 */ 12 class Solution { 13 func rotateRight(_ head: ListNode?, _ k: Int) -> ListNode? { 14 /* 15 * 思路,首先我们计算出链表的长度,这样据k值可以计算出链表实际移动距离 16 * 然后,通过两个指针来记录移动位置 17 */ 18 var length = 0 19 var temp = head 20 while temp?.next != nil { 21 length += 1 22 temp = temp?.next 23 } 24 25 // 长度已经获取length+1 26 let stride = k % (length + 1) 27 if stride == 0{ 28 return head 29 } 30 // stride为实际需要移动的距离 31 // 定义两个指针,一个指向链表倒数第stride+1个,一个指向最后一个 32 var start: ListNode? 33 var end: ListNode? 34 var tem: ListNode? 35 for index in 0...length{ 36 if index == 0 { 37 tem = head 38 end = head 39 }else{ 40 tem = tem?.next 41 end = end?.next 42 } 43 if index == length + 1 - (stride + 1) { 44 start = tem 45 } 46 } 47 48 /* 49 * 修改节点 50 */ 51 let res = start?.next 52 start?.next = nil 53 end?.next = head 54 55 return res 56 } 57 } 48ms 1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * public var val: Int 5 * public var next: ListNode? 6 * public init(_ val: Int) { 7 * self.val = val 8 * self.next = nil 9 * } 10 * } 11 */ 12 class Solution { 13 func rotateRight(_ head: ListNode?, _ k: Int) -> ListNode? { 14 guard let head = head else { 15 return nil 16 } 17 18 guard k > 0 else { 19 return head 20 } 21 22 var length = 0 23 var next: ListNode? = head 24 var tail = head 25 while next != nil { 26 tail = next! 27 next = next?.next 28 length += 1 29 } 30 31 if (length == 1) { 32 return head 33 } 34 35 let k = k > length ? k % length : k 36 if (k == length || k == 0) { 37 return head 38 } 39 40 next = head 41 for _ in 0..<length - k - 1 { 42 next = next?.next 43 } 44 let start = next?.next 45 tail.next = head 46 next?.next = nil 47 return start 48 } 49 } |
请发表评论