
问题
给定两个句子 A 和 B 。 (句子是一串由空格分隔的单词。每个单词仅由小写字母组成。)
如果一个单词在其中一个句子中只出现一次,在另一个句子中却没有出现,那么这个单词就是不常见的。
返回所有不常用单词的列表。
您可以按任何顺序返回列表。
输入:A = “this apple is sweet”, B = “this apple is sour”
输出:[“sweet”,”sour”]
输入:A = “apple apple”, B = “banana”
输出:[“banana”]
提示:
0 <= A.length <= 200
0 <= B.length <= 200
A 和 B 都只包含空格和小写字母。
We are given two sentences A and B. (A sentence is a string of space separated words. Each word consists only of lowercase letters.)
A word is uncommon if it appears exactly once in one of the sentences, and does not appear in the other sentence.
Return a list of all uncommon words.
You may return the list in any order.
Input: A = “this apple is sweet”, B = “this apple is sour”
Output: [“sweet”,”sour”]
Input: A = “apple apple”, B = “banana”
Output: [“banana”]
Note:
0 <= A.length <= 200
0 <= B.length <= 200
A and B both contain only spaces and lowercase letters.
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: 。
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