Algorithem_MergeTwoSortedLists

2022-04-22

Algorithem_MergeTwoSortedLists

You are given the heads of two sorted linked lists list1 and list2.

Merge the two lists in a one sorted list. The list should be made by splicing together the nodes of the first two lists.

Return the head of the merged linked list.

Example 1:

1 2	Input: list1 = [1,2,4], list2 = [1,3,4] Output: [1,1,2,3,4,4]

Example 2:

1 2	Input: list1 = [], list2 = [] Output: []

Example 3:

1 2	Input: list1 = [], list2 = [0] Output: [0]

解法

需求是合并两个有序的LinkList，所以解法是

判断 list1.val 和 list2.val 的大小，
如果 list1.val 小于等于 list2.val，则第一个元素是 list1.val，然后递归 list1.next 和 list2——即比较 list1.next.val 和 list2.val 的大小
如果 list1.val 大于 list2.val，则第一个元素是 list2.val，然后递归 list1和 list2.next——即比较 list1.val 和 list2.next.val 的大小
如果 list1为空，则返回 list2，如果 list2为空则返回 list1

代码如下：


/**
 * 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(_ list1: ListNode?, _ list2: ListNode?) -> ListNode? { 
        if list1 == nil {
            return list2
        }
        
        if list2 == nil {
            return list1
        }
        
        if list1!.val <= list2!.val {
            list1?.next = mergeTwoLists(list1?.next, list2)
            return list1
        }
        else {
            list2?.next = mergeTwoLists(list1, list2?.next)
            return list2
        }
    }
}