[Swift]LeetCode328. 奇偶链表 | Odd Even Linked List_app

概述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 pl

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: Output: 1->2->3->4->5->NulL1->3->5->2->4->NulL

Example 2:

input: 2Output: ->1->3->5->6->4->7->NulL2->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 = true10         while node != nil {11             node?.next = node?.next?.next12             if isOdd {13                 oddEnd = node14             }15             node = next16             next = next?.next17             isOdd = !isOdd18         }19         oddEnd?.next = evenhead20         return dummy.next21     }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? = head15         var even: ListNode? = head?.next16         var evenFirst: ListNode? = even17         18         while true {19             if odd == nil || even == nil || even?.next == nil {20                 odd?.next = evenFirst21                 break22             }23             24             // Connecting odd25             odd?.next = even?.next26             odd = even?.next27             28             if odd?.next == nil {29                 even?.next = nil30                 odd?.next = evenFirst31                 break32             }33             34             // Connection even35             even?.next = odd?.next36             even = even?.next37             38         }39         40         return head41     }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?.next15         var evennode = head?.next16         let evenheadNode = head?.next17         var node = head18         while oddNode != nil {19             evennode?.next = oddNode?.next20             node?.next = oddNode21             oddNode?.next = evenheadNode22             node = oddNode23             evennode = evennode?.next24             oddNode = evennode?.next25         }26         27         return head28     }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 = curOdd10         while cur?.next != nil  && curOdd?.next != nil{11             if index % 2 == 1 {12                 cur?.next = curOdd?.next13                 cur = cur?.next14             }else {15                 curOdd?.next = cur?.next16                 curOdd = curOdd?.next17             }18             index += 119         }20         cur?.next = dummyOdd.next21         curOdd?.next = nil22         return dummy.next23     }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 = head15         var j = head?.next16         let jH = j17         18         while i?.next?.next != nil || j?.next?.next != nil {19             if i?.next?.next != nil {20                 i?.next = i?.next?.next21                 i = i?.next22             }23             if j?.next?.next != nil {24                 j?.next = j?.next?.next25                 j = j?.next26             }27         }28         j?.next = nil // 清空偶数节点的next29         i?.next = jH30         return head31     }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 = head15         var second = head?.next16         let doulbeNode = second17         var lastCNode = head18         while first != nil || second != nil {19             first?.next = second?.next20             lastCNode = first21             first = first?.next22             second?.next = first?.next23             second = second?.next24         }25         lastCNode?.next = doulbeNode26         return head27     }28 }