LeetCode-771-Jewels and Stones(宝石和石头)

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本文介绍了一个简单的编程问题，即计算给定字符串S中有多少字符出现在另一字符串J中，这里的J代表宝石，而S代表普通的石头。文章提供了一段Java代码示例来解决这个问题。

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Q：

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.

Analysis:

此题的题意为给定字符串J为含有宝石的字符串，J中每个字符都可以成为一个宝石，S字符串是普通给定的石头，要查找出S中含有宝石的个数，即含有J中字符的个数，提示中说明J中字符均不相同，同时区分大小写。

解题思路：

直接遍历字符串S中的每个字符，同时判断如果该字符在J中出现，则S中含有宝石的数量+1，遍历完S后，即可得到结果，并将结果返回。

Code:

class Solution {
    public int numJewelsInStones(String J, String S) {
//      统计出现的次数
        int count=0;
//      将S遍历每个字符，查看String J中是否含有S的字符
        for(int i=0;i<S.length();i++){
            char c=S.charAt(i);
            String c1=String.valueOf(c);
            if(J.contains(c1)){
//              含有的话数量+1
                count++;
            }
        }
        return count;
    }
}