Jewels and Stones

題目

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.

翻譯：

簡單講就是，給你一個代表不同寶石的字串，和一個代表石頭的字串，從石頭字串裡找出屬於寶石的石頭，並返回個數。
注意，區分大小寫即a和A不一樣。

解答：

1.通過遍歷字串S和J，兩兩進行比較，判斷stone中有多少顆jewel。

• 時間複雜度為O(s * j)。(s為字串S的長度，j為字串J的長度)。
• 空間複雜度為O(1)。

java:

class Solution {
    public int numJewelsInStones(String J, String S) {
        int res = 0;
        Set set = new HashSet();
        for(char c:J.toCharArray())set.add(c);
        for(char s:S.toCharArray())if(set.contains(s))res++; 
        return
 
 res;
    }  
}

class Solution {
public:
    int numJewelsInStones(string J, string S) {
        int res = 0;
        set<char> setJ(J.begin(), J.end());
        for (char s : S) if (setJ.count(s)) res++;
        return res;       
    }
};

c++版本：

class Solution {
public:
    int numJewelsInStones(string J, string S) {
        int ret =0;
        set<char>set;
        for(int i = 0;i<J.length();i++)
            set.insert(J[i]);
        for(int i = 0;i<S.length();i++)
            if(set.count(S[i]))
                ret++;

        return ret;
    }
};