
问题
给定字符串J 代表石头中宝石的类型,和字符串 S代表你拥有的石头。 S 中每个字符代表了一种你拥有的石头的类型,你想知道你拥有的石头中有多少是宝石。
J 中的字母不重复,J 和 S中的所有字符都是字母。字母区分大小写,因此”a”和”A”是不同类型的石头。
输入: J = “aA”, S = “aAAbbbb”
输出: 3
输入: J = “z”, S = “ZZ”
输出: 0
注意:
S 和 J 最多含有50个字母。
J 中的字符不重复。
You’re given strings J representing the types of stones that are jewels, and S representing the stones you have. Each character in S is a type of stone you have. You want to know how many of the stones you have are also jewels.
The letters in J are guaranteed distinct, and all characters in J and S are letters. Letters are case sensitive, so “a” is considered a different type of stone from “A”.
Input: J = “aA”, S = “aAAbbbb”
Output: 3
Input: J = “z”, S = “ZZ”
Output: 0
Note:
S and J will consist of letters and have length at most 50.
The characters in J are distinct.
示例
public class Program { public static void Main(string[] args) { var J = "aA"; var S = "aAAbbbb"; var res = NumJewelsInStones(J, S); Console.WriteLine(res); J = "z"; S = "ZZ"; res = NumJewelsInStones2(J, S); Console.WriteLine(res); Console.ReadKey(); } private static int NumJewelsInStones(string J, string S) { var res = 0; foreach(var c in S) { if(J.Contains(c)) res++; } return res; } private static int NumJewelsInStones2(string J, string S) { var res = 0; var set = new HashSet<char>(); foreach(var c in J) { set.Add(c); } foreach(var c in S) { if(set.Contains(c)) res++; } return res; } }
以上给出2种算法实现,以下是这个案例的输出结果:
3 0
分析:
设石头的类型数量为 m,拥有的石头的数量为 n ,那么NumJewelsInStones 的时间复杂应当为: ,NumJewelsInStones2的时间复杂为:
。
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