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.

给两个字符串，珠宝字符串J和石头字符串S，其中J中的每个字符都是珠宝，S中的每个字符都是石头，区分字母的大小写，问我们S中有多少个珠宝。

Java:

public int numJewelsInStones(String J, String S) {

    return S.replaceAll("[^" + J + "]", "").length();

}

Java:

public int numJewelsInStones(String J, String S) {

        int res = 0;

        Set setJ = new HashSet();

        for (char j: J.toCharArray()) setJ.add(j);

        for (char s: S.toCharArray()) if (setJ.contains(s)) res++;

        return res;

    }

Python:

def numJewelsInStones(self, J, S):

    return sum(map(J.count, S))

Python:

def numJewelsInStones(self, J, S):

    return sum(map(S.count, J))

Python:　　

def numJewelsInStones(self, J, S):

    return sum(s in J for s in S)

Python:

def numJewelsInStones(self, J, S):

        setJ = set(J)

        return sum(s in setJ for s in S)

Python:

# Time:  O(m + n)

# Space: O(n)

class Solution(object):

    def numJewelsInStones(self, J, S):

        """

        :type J: str

        :type S: str

        :rtype: int

        """

        lookup = set(J)

        return sum(s in lookup for s in S)

Python: wo

class Solution(object):

    def numJewelsInStones(self, J, S):

        """

        :type J: str

        :type S: str

        :rtype: int

        """

        res = 0

        for i in S:

            if i in J:

                res += 1

        return res

C++:

int numJewelsInStones(string J, string S) {

        int res = 0;

        unordered_set<char> setJ(J.begin(), J.end());

        for (char s : S) if (setJ.count(s)) res++;

        return res;

    }

All LeetCode Questions List 题目汇总

巴特西

[LeetCode] 771. Jewels and Stones 珠宝和石头

All LeetCode Questions List 题目汇总

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