Swift 合并两个有序链表 - LeetCode

LeetCode

题目： 合并两个有序链表

将两个有序链表合并为一个新的有序链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。

示例:

代码语言：javascript复制

输入：1->2->4, 1->3->4
输出：1->1->2->3->4->4

代码一 迭代：

代码语言：javascript复制

/**
 * 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 mergeTwoLists(_ l1: ListNode?, _ l2: ListNode?) -> ListNode? {
    if l1 == nil {
        return l2
    }
    
    if l2 == nil {
        return l1
    }
    var newNode: ListNode? = nil
    var newNode1 = l1
    var newNode2 = l2
    if newNode1!.val < newNode2!.val {
        newNode = newNode1
        newNode1 = newNode1?.next
    } else {
        newNode = newNode2
        newNode2 = newNode2?.next
    }
    
    var tempNode = newNode
    while (newNode1 != nil) && (newNode2 != nil) {
        if newNode1!.val <= newNode2!.val {
            tempNode?.next = newNode1
            newNode1 = newNode1?.next
        } else {
            tempNode?.next = newNode2
            newNode2 = newNode2?.next
        }
        tempNode = tempNode?.next
    }
    
    if newNode1 != nil && newNode2 == nil {
        tempNode?.next = newNode1
    }
    
    if newNode1 == nil && newNode2 != nil {
        tempNode?.next = newNode2
    }
    
    return newNode
  }
}

代码二 递归：

代码语言：javascript复制

/**
 * 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 mergeTwoLists(_ l1: ListNode?, _ l2: ListNode?) -> ListNode? {
          if l1 == nil && l2 == nil {
        return nil
    }
    
    if l1 == nil || l2 == nil {
        return l1 == nil ? l2 : l1
    }
    
    var newNode: ListNode? = nil
    
    if l1!.val <= l2!.val {
        newNode = l1
        newNode?.next = mergeTwoLists(l1?.next, l2)
    } else {
        newNode = l2
        newNode?.next = mergeTwoLists(l1, l2?.next)
    }
    return newNode  
  }
}

swift

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