LeetCode刷题：771. Jewels and Stones

LeetCode刷题：771. Jewels and Stones

原题链接：https://leetcode.com/problems/jewels-and-stones/

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.

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。

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。

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。

算法设计

package com.bean.algorithm.basic;public class JewelsandStones {/** Input: J = "aA", S = "aAAbbbb"* Output: 3* */public int numJewelsInStones(String J, String S) {char[] jewels = J.toCharArray();char[] stones = S.toCharArray();int found = 0;for(int i = 0; i < stones.length; i++) {for(int j = 0; j < jewels.length; j++) {if(jewels[j] == stones[i]) {found++;}}}return found;}public static void main(String[] args) {// TODO Auto-generated method stubJewelsandStones jewelsandstones=new JewelsandStones();String J= "aA";String S="aAAbbbb";int result = jewelsandstones.numJewelsInStones(J, S);System.out.println("result = "+result);}}

程序运行结果：

result = 3

LeetCode上提交代码，Accepted。