771. Jewels and Stones

leetcode

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.

先说思路

J里面的元素一定是不同的，所以我就想，从J里面每次取一个元素，然后和S的每一个元素进行比较，相同就加一，然后进行下一个元素的比较，完美，没毛病。

int numJewelsInStones(char* J, char* S){
    int count=0;
    int len1=strlen(J);
    int len2=strlen(S);
    for(int i=0;i<len1;i++){
        for(int j=0;j<len2;j++){
            if(J[i]==S[j]){
                count+=1;
            }
        }
    }
    return count;
}

Runtime: 4 ms, faster than 100.00% of C online submissions for Jewels and Stones.
真的easy