771. Jewels and Stones - LeetCode

Question

771. Jewels and Stones

Solution

题目大意:两个字符串J和S,其中J中每个字符不同,求S中包含有J中字符的个数,重复的也算

思路:Set记录字符串J中的每个字符,遍历S中的字符,如果出现在Set中,count加1

Java实现:

public int numJewelsInStones(String J, String S) {
    Set<Character> set = new HashSet<>();
    int count = 0;
    for (char c : J.toCharArray()) {
        set.add(c);
    }
    for (char c : S.toCharArray()) {
        if (set.contains(c)) {
            count++;
        }
    }
    return count;
}

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转载自www.cnblogs.com/okokabcd/p/9368223.html