LeetCode 771. Jewels and Stones--Java和Python解法--简单

博客围绕计算拥有的宝石中珠宝数量的问题展开,给定代表珠宝类型的字符串J和代表拥有宝石的字符串S,需计算S中有多少宝石也是珠宝,同时提到字母区分大小写,还给出了Python和Java代码示例。

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题目地址:Jewels and Stones - LeetCode



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”.

Example 1:

Input: J = "aA", S = "aAAbbbb"
Output: 3

Example 2:

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.

你会得到J代表珠宝类型的字符串,S代表你拥有的宝石。每个角色S都是你拥有的一种石头。你想知道你有多少宝石也是珠宝。

在这些信件J是保证不同,而在所有的字符J和S是字母。字母区分大小写,因此"a"被认为是不同类型的石头"A"。


python代码如下:

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

Java代码如下:

class Solution {
    public int numJewelsInStones(String J, String S) {
        int count = 0;
        for (char s: S.toCharArray()) {
            if (J.contains(""+s)){
             count+=1;
            }
        }
        return count;
    }
}
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