你得到的字符串J代表珠宝的类型，S代表你拥有的宝石。 S中的每个分子都是你拥有的一种宝石。你想知道你有多少宝石是规定类型的珠宝。
J中的字母保证每个都不同，J和S中的所有字符都是字母。字母区分大小写，因此“a”被认为是与“A”不同类型的宝石。Example 1:
Input: J = "aA", S = "aAAbbbb"
Output: 3
Example 2:
Input: J = "z", S = "ZZ"
Output: 0注：S和J将由字母组成，长度最多为50。
J中的字符是截然不同的。解体思路：
其实这题就是一个字符串J，一个字符串S，在S中寻找J字符串中每个字节出现的次数。
所以我们构建两层循环，第一层将字符串J拆成一个个char，第二层将字符串S拆成一个个char，然后将字符串J中的每个char去字符串S中寻找一次。
由于备注中说明J中无重复的字符，所以不用特殊处理，每次找到给总数+1就够了。
package cn.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.
*
* @author kimtian
*/
public class JewelsAndStones {
public static int numJewelsInStones(String J, String S) {
int sum = 0;
// 字符串S不在规定的范围中,则返回0
if (S.length() > 50 && S.length() <= 0) {
return 0;
} else {
//先遍历J字符串
for (int i = 0; i < J.length(); i++) {
char j = J.charAt(i);
//再遍历S字符串
for (int z = 0; z < S.length(); z++) {
//如果中是否包含这个char，则总数加1
if (S.charAt(z) == j) {
sum++;
}
}
}
}
return sum;
}
/**
* 测试类
*
* @param args
*/
public static void main(String[] args) {
//两个字符串都为空
System.out.println(numJewelsInStones("", ""));
//结果应该为3
System.out.println(numJewelsInStones("Aa", "aAAbbbb"));
//结果应该为2
System.out.println(numJewelsInStones("A", "aAAbbbb"));
//结果应该为1
System.out.println(numJewelsInStones("a", "aAAbbbb"));
//因为区分大小写，所以结果应该为0
System.out.println(numJewelsInStones("B", "aAAbbbb"));
}
}