LeetCode-771. Jewels and Stones

最新推荐文章于 2025-06-25 07:15:00 发布

转载最新推荐文章于 2025-06-25 07:15:00 发布 · 61 阅读

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原文链接：http://www.cnblogs.com/zhacai/p/10429670.html

本文介绍了一种算法，用于计算给定字符串中属于特定类型字符（宝石）的数量。通过使用Map或Set数据结构，该算法能高效地遍历并计数输入字符串中符合宝石类型的字符。示例代码展示了两种实现方式，一种利用Map存储宝石类型及其出现次数，另一种使用Set快速检查字符是否为宝石。

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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.

public int numJewelsInStones(String J, String S) {//my
        Map<Character,Integer> map = new HashMap<Character, Integer>();
        int re=0;
        for (int i = 0; i < J.length(); i++) {
            map.put(J.charAt(i),0);
        }
        for (int i = 0; i < S.length(); i++) {
            if(map.containsKey(S.charAt(i))){
                map.put(S.charAt(i),map.get(S.charAt(i))+1);
            }
        }
        for (Map.Entry entry:map.entrySet()) {
            re += (Integer) entry.getValue();
        }
        return  re;
    }

简洁版，使用Set

public int numJewelsInStones(String J, String S) {//my
        int re=0;
        Set<Character> set = new HashSet<Character>();
        for (int i = 0; i < J.length(); i++) {
            set.add(J.charAt(i));
        }
        for (int i = 0; i < S.length(); i++) {
            if(set.contains(S.charAt(i))){
                re++;
            }
        }
        return  re;
    }

转载于:https://www.cnblogs.com/zhacai/p/10429670.html