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

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;
    }

　　

　　

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