LeetCode 771 Jewels and Stones
（本人专注于刷leetcode全部题c#语言算法，后期发视频教程）
相信有些人看到这道题目的时候不会一下子明白。
首先解释一下题意：
理解/核心： J 中字符 在 S 中出现的次数，区分大小写。
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: 解释:a字母出现在S的字符串中的次数为1，A 出现在S字符串的次数为2 . 1+2=3的问题！！！！
Input: J = "aA", S = "aAAbbbb" Output: 3
Example 2: 解释：z在S中出现了0次，结果为0，原因：区分大小写。（读懂了题目就简单了）
Input: J = "z", S = "ZZ" Output: 0
c#版本答案：（未考虑时间复杂度，较为理解）
static void Main(string[] args)
{
string J = "zZsu";
string P = "zZZasuidassssssss";
int pc = GetNum(J, P);
Console.WriteLine(pc);
Console.ReadKey();
}
public static int GetNum(string J,string P)
{
int num = 0;
for (int i = 0; i < P.Length; i++)
{
for (int j = 0; j < J.Length; j++)
{
if (J[j] == P[i])
{
num++;
}
}
}
return num;
}
c#版本答案：（考虑空间复杂度的算法）
int[]count =new int[64];
char[] ooo= J.ToCharArray();
for (char c=' ' ; c < J.ToCharArray().Length; c++)
{
count[c - 'A']++;
}
int ans = 0;
for (char c =' '; c < P.ToCharArray().Length; c++)
{
if (count[c - 'A'] >= 1)
ans++;
}
return ans; 如果对答案有异议，欢迎各路大神指点，并在评论区留下c#代码，相互学习。