[LeetCode]771. 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.

代码

int numJewelsInStones(char* J, char* S) {
    int count = 0;
    int jlen=strlen(J);
    int slen=strlen(S);
    for(int i=0;i<jlen;i++){
        for(int j=0;j<slen;j++){
            if(S[j]==J[i]){
                count++;
            }
        }
    }
    return count;
}

个人

开始时我按照note中给的长度不超多50来计算，导致结束符等被计入，输出结果错误，应该先利用函数得出长度，进而比较计数。