题目:
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.
Answer:
我的答案:
class Solution:
def numJewelsInStones(self, J, S):
"""
:type J: str
:type S: str
:rtype: int
"""
count = 0
for m in J:
for n in S:
if m == n:
count += 1
return count
其它答案用到map()和string.count():
def numJewelsInStones(self, J, S):
return sum(map(J.count, S))
def numJewelsInStones(self, J, S):
return sum(map(S.count, J)) # this one after seeing https://discuss.leetcode.com/post/244105
def numJewelsInStones(self, J, S):
return sum(s in J for s in S)