原問題：

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".

Note: • S and J will consist of letters and have length at most 50. • The characters in J are distinct.

直接查詢求解：

//max_n1為J陣列的最大長度，max_n2為S陣列的最大長度

int JewelsNum_1(const char *J,int max_n1,const char *S,int max_n2)
{
    int i = 0,j,num=0;
    while(S[i]!='\0' && i<max_n2)
    {
        j = 0;
        while(J[j]!='\0' && j<max_n1)
        {
            if(S[i] == J[j])
            {
                num++;
                break;
            }
            j++;
        }
        i++;
    }
    return num;
}
 

雜湊查詢求解：

int JewelsNum_2(const char *J,int max_n1,const char *S,int max_n2)
{
    int i = 0,num=0,all[52] = {
        0
    };
    char temp = '\0';
    temp = S[i];
    while(temp!='\0' && i<max_n2)
    {
        if(temp>='a' && temp<='z')
        {
            all[temp-'a']++;
        }
        else if(temp>='A' && temp<='Z')
        {
            all[temp-'A'+26]++;
        }
        else;
        i++;
        temp = S[i];
    }
    i = 0;
    temp = J[i];
    while(temp!='\0' && i<max_n1)
    {
        if(temp>='a' && temp<='z')
        {
            num = num + all[temp-'a'];
        }
        else if(temp>='A' && temp<='Z')
        {
            num = num + all[temp-'A'+26];
        }
        else;
        i++;
        temp = J[i];
    }
    return num;
}
 

總結：

直接查詢演算法最壞情況下的時間複雜度為m*n，雜湊演算法最壞情況下的時間複雜度為m+n。