[Swift]LeetCode86. 分隔链表 | Partition List

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Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.

You should preserve the original relative order of the nodes in each of the two partitions.

Example:

Input: head = 1->4->3->2->5->2, x = 3
Output: 1->2->2->4->3->5

---

给定一个链表和一个特定值 x，对链表进行分隔，使得所有小于 x 的节点都在大于或等于 x 的节点之前。

你应当保留两个分区中每个节点的初始相对位置。

示例:

输入: head = 1->4->3->2->5->2, x = 3
输出: 1->2->2->4->3->5

---

16ms

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public var val: Int
 *     public var next: ListNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.next = nil
 *     }
 * }
 */
class Solution {
    func partition(_ head: ListNode?, _ x: Int) -> ListNode? {
        var head = head
        
        var leftHead = ListNode(0)
        let leftDummy = leftHead
        
        var rightHead = ListNode(0)
        var rightDummy = rightHead
        
        while head != nil {
            if head!.val < x {
                leftHead.next = head
                leftHead = leftHead.next ?? ListNode(0)
            }else {
                rightHead.next = head
                rightHead = rightHead.next ?? ListNode(0)
            }
            head = head!.next
        }
        
        leftHead.next = rightDummy.next
        rightHead.next = nil
        return leftDummy.next
    }
}

---

16ms

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public var val: Int
 *     public var next: ListNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.next = nil
 *     }
 * }
 */
class Solution {
    func partition(_ head: ListNode?, _ x: Int) -> ListNode? {
        let prevDummy = ListNode(0), postDummy = ListNode(0)
        var prev = prevDummy, post = postDummy
        
        var node = head
        
        while node != nil {
            let next = node!.next
            node!.next = nil

            if node!.val < x {
                prev.next = node
                prev = prev.next!
            } else {
                post.next = node
                post = post.next!
            }
            node = next
        }
        
        prev.next = postDummy.next
        
        return prevDummy.next
    }
}

---

20ms

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public var val: Int
 *     public var next: ListNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.next = nil
 *     }
 * }
 */
class Solution {
    func partition(_ head: ListNode?, _ x: Int) -> ListNode? {
        let dummyHead1 = ListNode(0)
        let dummyHead2 = ListNode(0)
        var p1 = dummyHead1
        var p2 = dummyHead2
        var p = head
        while p != nil {
            if p!.val < x {
                p1.next = p
                p1 = p!
            } else {
                p2.next = p
                p2 = p!
            }
            p = p!.next
        }
        p1.next = dummyHead2.next
        p2.next = nil
        return dummyHead1.next
    }
}