内容简介：You are 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 guaran
You are 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.
return J.reduce(0) { sum, j in
sum + S.filter({$0 == j}).count
}
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深度探索C++对象模型
[美] Stanley B. Lippman / 侯捷 / 华中科技大学出版社 / 2001-5 / 54.00元
这本书探索“对象导向程序所支持的C++对象模型”下的程序行为。对于“对象导向性质之基础实现技术”以及“各种性质背后的隐含利益交换”提供一个清楚的认识。检验由程序变形所带来的效率冲击。提供丰富的程序范例、图片，以及对象导向观念和底层对象模型之间的效率测量。一起来看看 《深度探索C++对象模型》 这本书的介绍吧!
