174.Jewels and Stones

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.

思路

问题抽象为：找S中包含有几个J中的字符；
1.取出J中所有不同的字符全部放到set中；
2.遍历S中的字符，如果在map1中则num++；
3.return num

代码

java

public int numJewelsInStones(String J, String S) {int num = 0;Set<Character> set = new HashSet<Character>(); for (int i = 0; i< J.length(); i++) {set.add(J.charAt(i));}for (int i = 0; i< S.length(); i++) {if (set.contains(S.charAt(i))) {num++;}}return num;}

python

def numJewelsInStones(self, J, S):""":type J: str:type S: str:rtype: int"""#定义一个set，存放J中不同的字符s1 = set()num = 0for j in J : s1.add(j)for s in S :if s in s1 :num = num + 1return num

C++

 int numJewelsInStones(string J, string S) {int num = 0;std::set<char> s1;for (int i = 0; i < J.size(); i++) {s1.insert(J[i]);}for (int i = 0; i< S.size(); i++) {if (s1.find(S[i]) != s1.end()) {num++;}}return num;}

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174.Jewels and Stones

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