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
andJ
will consist of letters and have length at most 50.- The characters in
J
are distinct.
翻译:
你被给定字符串 J
表示石头类型是珠宝, S
表示你有的石头。每一个 S
中的字符是一种你拥有的石头。你想知道你拥有的多少石头同时也是珠宝。
J
中的字母保证是不同的,并且 J
和 S
中的所有字符都是字母。字母大小写敏感,因此,"a"
被看作是和 "A"
不同的。
例子1:
输入: J = "aA", S = "aAAbbbb" 输出: 3
例子2:
输入: J = "z", S = "ZZ" 输出: 0
注意:
S
和J
包含的字符长度不超过50。J
中的字符是不同的。
思路:
用set存储 J 中不同的字符。然后扫描 S 中的字符看是否在set中,若在,则说明是珠宝,结果result++。
C++代码(Visual Studio2017):
#include "stdafx.h" #include <iostream> #include <string> #include <set> using namespace std; class Solution { public: int numJewelsInStones(string J, string S) { int result=0; set<char> setJ(J.begin(), J.end()); for (char s : S) if (setJ.count(s)) result++; return result; } }; int main() { Solution s; string J = "aA"; string S = "aAAbbbb"; int result; result = s.numJewelsInStones(J, S); cout << result; return 0; }