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algortthm

admin 11月 14, 2020 0

Jewels and Stones

Title Detail

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:

1 2	Input: J = "aA", S = "aAAbbbb" Output: 3

Example 2:

1 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

使用Java里面的 Map。key键保存宝石类型，value键保存数量统计。

Algorithm

class  {
    public int numJewelsInStones(String J, String S) {
        Map<Character, Integer> map = new HashMap<Character, Integer>();
        for(int i=0; i<J.length(); i++){
            map.put(J.charAt(i), 0);
        }
        for(int i=0; i<S.length();i++){
            Character c = S.charAt(i);
            if(map.get(c)!=null){
                Integer freq =map.get(c);
                map.put(c, freq+1);
            }
        }
        int result = 0;
        for(Integer value: map.values()){
            result += value;
        }
        return result;
    }
}

法2 正则表达式

1
2
3

String  a = "[^"+J+"]";
        S = S.replaceAll(a, "");
        return S.length();

Python方法

1	return sum(S.count(i) for i in J)