8/29 每日一題(檢查石頭中有多少珠寶,區分大小寫)

 You're given strings jewels representing the types of stones that are jewels, and stones representing the stones you have. Each character in stones is a type of stone you have. You want to know how many of the stones you have are also jewels.

Letters are case sensitive, so "a" is considered a different type of stone from "A".

 

Example 1:

Input: jewels = "aA", stones = "aAAbbbb"
Output: 3

Example 2:

Input: jewels = "z", stones = "ZZ"
Output: 0

 

Constraints:

• 1 <= jewels.length, stones.length <= 50
• jewels and stones consist of only English letters.
• All the characters of jewels are unique.

class Solution:

    def numJewelsInStones(self, jewels: str, stones: str) -> int:

        res=0

        j_list=list(jewels)

        for i in j_list:

            res+=stones.count(i)

        return res

---------------

from collections import defaultdict
class Solution:
    def numJewelsInStones(self, jewels: str, stones: str) -> int:
        jew = set()
        stCount = 0
        for i in range(0, len(jewels)):
            jew.add(jewels[i])
        for i in range(0,len(stones)):
            if stones[i] in jew:
                stCount += 1
        return stCount

----------------

class Solution:
    def numJewelsInStones(self, jewels: str, stones: str) -> int:
        l=len(jewels)
        l2=len(stones)
        c=0
        for i in range(0,l):
            for j in range(0,l2):
                if jewels[i]==stones[j]:
                    c+=1
        return c
-----------------------------
class Solution:
    def numJewelsInStones(self, jewels: str, stones: str) -> int:
        d={}
        for i in jewels:
            if i in d:
                d[i]+=1
            else:
                d[i]=1
        ans=0
        for j in stones:
            if j in d:
                ans+=d[j]
        return ans