[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:

?

1 2 3

public int numJewelsInStones(String J, String S) {     return S.replaceAll("[^" + J + "]", "").length(); }

Java:

?

1 2 3 4 5 6 7

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:

?

1 2	def numJewelsInStones(self, J, S):     return sum(map(J.count, S))

Python:

?

1 2	def numJewelsInStones(self, J, S):     return sum(map(S.count, J))

Python:　　

?

1 2	def numJewelsInStones(self, J, S):     return sum(s in J for s in S)

Python:

?

1 2 3

def numJewelsInStones(self, J, S):         setJ = set(J)         return sum(s in setJ for s in S)

Python:

?

1 2 3 4 5 6 7 8 9 10 11

# 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

?

1 2 3 4 5 6 7 8 9 10 11 12 13

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++:

?

1 2 3 4 5 6

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