leetcode 771 solution

最新推荐文章于 2020-10-20 09:18:15 发布

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最新推荐文章于 2020-10-20 09:18:15 发布

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本文介绍了一种算法，用于计算给定的石头中，有多少是特定类型的宝石。通过双层循环遍历，对比每个字符来判断石头是否为宝石，最终返回匹配的数量。示例展示了当输入J=aA S=aAAbbbb时，输出为3；当输入J=z S=ZZ时，输出为0。

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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

Solution:

class Solution {
    public int numJewelsInStones(String J, String S) {
        int result = 0;
        for(int i = 0; i<J.length(); ++i){
            for(int j = 0; j < S.length(); ++j){
                char a = J.charAt(i);
                char b = S.charAt(j);
                if(a == b){
                    result++;
                }
                
            }
        }
        return result;
        
        
    }
}