LeetCode 771: 寶石與石頭 Jewels and Stones

2019 年 11 月 20 日
筆記

題目:

給定字元串J 代表石頭中寶石的類型，和字元串 S代表你擁有的石頭。S 中每個字元代表了一種你擁有的石頭的類型，你想知道你擁有的石頭中有多少是寶石。

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.

J 中的字母不重複，J 和 S中的所有字元都是字母。字母區分大小寫，因此"a"和"A"是不同類型的石頭。

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".

示例 1:

輸入: J = "aA", S = "aAAbbbb"  輸出: 3

示例 2:

輸入: J = "z", S = "ZZ"  輸出: 0

注意:

S 和 J 最多含有50個字母。
J 中的字元不重複。

Note:

S and J will consist of letters and have length at most 50.
The characters in J are distinct.

解題思路:

J 改為 Set 集合, 遍歷 S 即可(因為 Set 查找複雜度為常數)

Java:

class Solution {      public int numJewelsInStones(String J, String S) {          Set<Character> set = new HashSet<>();          for (char c : J.toCharArray())              set.add(c);          int count = 0;          for (char c : S.toCharArray())              if (set.contains(c)) count++;          return count;      }  }

Python:

class Solution:      def numJewelsInStones(self, J: str, S: str) -> int:          count = 0          hash_set = set(J)          for c in S:              if c in hash_set:                  count += 1          return count