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
andJ
will consist of letters and have length at most 50.- The characters in
J
are distinct.
class Solution(object): def numJewelsInStones(self, J, S): """ :type J: str :type S: str :rtype: int out("", "")=0 out("", "A")=0 out("A", "")=0 out("AB", "C")=0 out("C", "AB")=0 out("AB", "AB")=0 out("ABC", "DEF")=0 out("ABC", "ABc")=2 out("ABC", "ABCD")=3 """ ''' if not J: return 0 jewels = set() for j in J: jewels.add(j) ans = 0 for s in S: if s in jewels: ans += 1 return ans ''' setJ = set(J) return sum(s in setJ for s in S)
时间复杂度O(S+J) 空间复杂度 O(S+J)
注: sum([True, False, True])==2