LeetCode 771.Jewels and Stones

Description

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 character in J are distinct.

My Submission

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

解题思路：

首先想到使用两层嵌套循环，外层为J，内层为S，遍历S中元素j依次和J的i个元素比较，若相等则计数器加一，最后返回计数器。

遇到问题：

刚开始不知道C字符串长度的计算方法，经查阅百度得知可以使用strlen()函数。同时有一个疑问，不清楚指针和数组名是否可以进行同等操作，现在想来应该是一样的。

Sample Submission

sample 0 ms submission：

int numJewelsInStones(char* J, char* S) {
    int result_num = 0;
    char* src = J;
    char* dst = S;
    

    int i,j;
    
    if ((J == NULL) || (S = NULL))
        return -1;
    
    for (i=0;i<50;i++)
    {
        if (src[i] == '\0')
            break;
        for (j=0;j<50;j++)
        {
            if (dst[j] == '\0')
                break;
            if (src[i] == dst[j]) {
                //printf("test %c:%c\n", src[i], dst[j]);
                result_num++;
            }
        }
    }
        
        
    return result_num;
}

sample 7880 kb submission：

int numJewelsInStones(char* J, char* S) {
    bool hash[53]={0};
    while(*J)
    {
        if(*J<='z' && *J>='a' )
            hash[(unsigned)((*J)-'a')]=true;
        else
            hash[(unsigned)((*J)-'A'+26)]=true;
        J++;
    }
    int count = 0;
    while(*S)
    {
        if(*S<='z' && *S>='a' && hash[(unsigned)((*S)-'a')])
            count++;
        if(*S<='Z' && *S>='A' && hash[(unsigned)((*S)-'A')+26])
            count++;
        S++;
    }
    return count;
}