LeetCode合并两个有序链表Swift

将两个升序链表合并为一个新的 升序 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：

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

暴力解法：

合并为一个链表

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public var val: Int
 *     public var next: ListNode?
 *     public init() { self.val = 0; self.next = nil; }
 *     public init(_ val: Int) { self.val = val; self.next = nil; }
 *     public init(_ val: Int, _ next: ListNode?) { self.val = val; self.next = next; }
 * }
 */
class Solution {
    func mergeTwoLists(_ l1: ListNode?, _ l2: ListNode?) -> ListNode? {
        var left = l1
        var right = l2
        
        let head = ListNode.init(-1)
        var current:ListNode? = head
        
        while left != nil && right != nil {
            if left!.val > right!.val {
                current?.next = right
                right = right?.next
            } else {
                current?.next = left
                left = left?.next
            } 
            current = current?.next
        }
        current?.next = (left == nil) ? right : left
        return head.next 
    }
}

递归解法：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public var val: Int
 *     public var next: ListNode?
 *     public init() { self.val = 0; self.next = nil; }
 *     public init(_ val: Int) { self.val = val; self.next = nil; }
 *     public init(_ val: Int, _ next: ListNode?) { self.val = val; self.next = next; }
 * }
 */
class Solution {
    func mergeTwoLists(_ l1: ListNode?, _ l2: ListNode?) -> ListNode? {
        guard let l1 = l1 else {
            return l2
        }
        
        guard let l2 = l2 else {
            return l1
        }
        
        if l1.val < l2.val {
            l1.next = mergeTwoLists(l1.next, l2)
            return l1
        } else {
            l2.next = mergeTwoLists(l1, l2.next)
            return l2
        }
    }
}