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.
解题思路:
J 用字典或者集合(Python 中判断集合中元素是否存在也是 O(1) 的时间)存储起来,然后遍历 S 中的每一个字符,判断字符是否也出现在 J 中。
时间复杂度 O(M+N)
Python 实现:
class Solution:
# AC: O(len(J) + len(S))
def numJewelsInStones(self, J, S):
"""
:type J: str
:type S: str
:rtype: int
"""
J = set(J) # Python 中 set 判断元素是否存在的时间复杂度为 O(1)
return sum(ch in J for ch in S)
a = "aA"
b = "aAAbbbb"
print(Solution().numJewelsInStones(a, b)) # 3
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