★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：https://www.cnblogs.com/strengthen/p/10260813.html
➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes.
You should try to do it in place. The program should run in O(1) space complexity and O(nodes) time complexity.
Example 1:
Input:1->2->3->4->5->NULLOutput:1->3->5->2->4->NULL
Example 2:
Input: 2->1->3->5->6->4->7->NULLOutput:2->3->6->7->1->5->4->NULL
Note:
- The relative order inside both the even and odd groups should remain as it was in the input.
- The first node is considered odd, the second node even and so on ...
给定一个单链表，把所有的奇数节点和偶数节点分别排在一起。请注意，这里的奇数节点和偶数节点指的是节点编号的奇偶性，而不是节点的值的奇偶性。
请尝试使用原地算法完成。你的算法的空间复杂度应为 O(1)，时间复杂度应为 O(nodes)，nodes 为节点总数。
示例 1:
输入: 1->2->3->4->5->NULL 输出: 1->3->5->2->4->NULL
示例 2:
输入: 2->1->3->5->6->4->7->NULL 输出: 2->3->6->7->1->5->4->NULL
说明:
- 应当保持奇数节点和偶数节点的相对顺序。
- 链表的第一个节点视为奇数节点，第二个节点视为偶数节点，以此类推。
36ms
1 class Solution { 2 func oddEvenList(_ head: ListNode?) -> ListNode? { 3 let dummy = ListNode(0) 4 var node = head 5 var next = head?.next 6 var oddEnd: ListNode? 7 let evenHead = head?.next 8 dummy.next = node 9 var isOdd = true 10 while node != nil { 11 node?.next = node?.next?.next 12 if isOdd { 13 oddEnd = node 14 } 15 node = next 16 next = next?.next 17 isOdd = !isOdd 18 } 19 oddEnd?.next = evenHead 20 return dummy.next 21 } 22 }
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 oddEvenList(_ head: ListNode?) -> ListNode? { 14 var odd: ListNode? = head 15 var even: ListNode? = head?.next 16 var evenFirst: ListNode? = even 17 18 while true { 19 if odd == nil || even == nil || even?.next == nil { 20 odd?.next = evenFirst 21 break 22 } 23 24 // Connecting odd 25 odd?.next = even?.next 26 odd = even?.next 27 28 if odd?.next == nil { 29 even?.next = nil 30 odd?.next = evenFirst 31 break 32 } 33 34 // Connection even 35 even?.next = odd?.next 36 even = even?.next 37 38 } 39 40 return head 41 } 42 }
44ms
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 oddEvenList(_ head: ListNode?) -> ListNode? { 14 var oddNode = head?.next?.next 15 var evenNode = head?.next 16 let evenHeadNode = head?.next 17 var node = head 18 while oddNode != nil { 19 evenNode?.next = oddNode?.next 20 node?.next = oddNode 21 oddNode?.next = evenHeadNode 22 node = oddNode 23 evenNode = evenNode?.next 24 oddNode = evenNode?.next 25 } 26 27 return head 28 } 29 }
56ms
1 class Solution { 2 func oddEvenList(_ head: ListNode?) -> ListNode? { 3 let dummy = ListNode(0) 4 let dummyOdd = ListNode(0) 5 dummy.next = head 6 var cur = head 7 var curOdd = head?.next 8 var index = 1 9 dummyOdd.next = curOdd 10 while cur?.next != nil && curOdd?.next != nil{ 11 if index % 2 == 1 { 12 cur?.next = curOdd?.next 13 cur = cur?.next 14 }else { 15 curOdd?.next = cur?.next 16 curOdd = curOdd?.next 17 } 18 index += 1 19 } 20 cur?.next = dummyOdd.next 21 curOdd?.next = nil 22 return dummy.next 23 } 24 }
64ms
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 oddEvenList(_ head: ListNode?) -> ListNode? { 14 var i = head 15 var j = head?.next 16 let jH = j 17 18 while i?.next?.next != nil || j?.next?.next != nil { 19 if i?.next?.next != nil { 20 i?.next = i?.next?.next 21 i = i?.next 22 } 23 if j?.next?.next != nil { 24 j?.next = j?.next?.next 25 j = j?.next 26 } 27 } 28 j?.next = nil // 清空偶数节点的next 29 i?.next = jH 30 return head 31 } 32 }
92ms
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 oddEvenList(_ head: ListNode?) -> ListNode? { 14 var first = head 15 var second = head?.next 16 let doulbeNode = second 17 var lastCNode = head 18 while first != nil || second != nil { 19 first?.next = second?.next 20 lastCNode = first 21 first = first?.next 22 second?.next = first?.next 23 second = second?.next 24 } 25 lastCNode?.next = doulbeNode 26 return head 27 } 28 }