C#LeetCode刷题之#21-合并两个有序链表（Merge Two Sorted Lists）

问题

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。 

输入：1->2->4, 1->3->4

输出：1->1->2->3->4->4

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

Input: 1->2->4, 1->3->4

Output: 1->1->2->3->4->4

示例

public class Program {

    public static void Main(string[] args) {
        var A = "this apple is sweet";
        var B = "this apple is sour";

        var res = UncommonFromSentences(A, B);
        ShowArray(res);

        Console.ReadKey();
    }

    private static void ShowArray(IList<string> array) {
        foreach(var domain in array) {
            Console.Write($"{domain} ");
        }
        Console.WriteLine();
    }

    private static string[] UncommonFromSentences(string A, string B) {
        string[] wordA = A.Split(' ');
        string[] wordB = B.Split(' ');
        var dicA = new Dictionary<string, int>();
        var dicB = new Dictionary<string, int>();
        var res = new List<string>();
        foreach(var word in wordA) {
            if(dicA.ContainsKey(word)) {
                dicA[word]++;
            } else {
                dicA[word] = 1;
            }
        }
        foreach(var word in wordB) {
            if(dicB.ContainsKey(word)) {
                dicB[word]++;
            } else {
                dicB[word] = 1;
            }
        }
        foreach(var kvp in dicA) {
            if(kvp.Value == 1 && !dicB.ContainsKey(kvp.Key)) {
                res.Add(kvp.Key);
            }
        }
        foreach(var kvp in dicB) {
            if(kvp.Value == 1 && !dicA.ContainsKey(kvp.Key)) {
                res.Add(kvp.Key);
            }
        }
        return res.ToArray();
    }

}

以上给出1种算法实现，以下是这个案例的输出结果：

sweet sour

分析：

显而易见，以上算法的时间复杂度为： 。