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.

 

J是没有重复字母的字符串，S是可能存在重复的字符串。计算S中所包含的S中和J中都存在的字母的个数。

对S进行排序再求解，减少在J中的遍历次数

代码如下：

class Solution {
public:
    int numJewelsInStones(string J, string S) {
        int res=0;
        sort(S.begin(),S.end());
        for(int i=0;i<S.size();++i){
            for(int j=0;j<J.size();++j){
                if(S[i]==J[j]){
                    res++;
                    while(S[i]==S[i+1]){
                        ++i;
                        ++res;
                    }
                }
            }
        }
    }
};