leetcode刷题记录 easy(1) 771.Jewels and Stones

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本文介绍了一个编程问题的解决方案，即如何计算给定石头集合中属于宝石的石头数量。通过使用Java中的indexOf方法，我们能够有效地遍历石头集合，判断每个石头是否属于宝石类型，并累加计数。

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英文题目：

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.

中文题目解释：

解析：

可以使用String类的indexOf方法，

public int indexOf(int ch): 返回指定字符在字符串中第一次出现处的索引，如果此字符串中没有这样的字符，则返回 -1。

结果提交：

class Solution {
    public int numJewelsInStones(String J, String S) {
        char[] sc=S.toCharArray();
        int i=0;
        for (char s:sc) {
                if(J.indexOf(s)!=-1){
                    i++;
                }
        }
        return i;
    }
}