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LeetCode 771: 宝石与石头 Jewels and Ston

IT图书 作者:karson 时间:2021-09-09 10:38:39 0 删除 编辑

题目:

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。

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。

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:

输入: J = "aA", S = "aAAbbbb"
输出: 3

示例 2:

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。

Note:

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

解题思路:

J 改为 Set 集合, 遍历 S 即可(因为 Set 查找复杂度为常数)

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

来自 “ ITPUB博客 ” ,链接:https://blog.itpub.net/1542/viewspace-2824471/ ,如需转载,请注明出处,否则将追究法律责任。

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