题目描述 LeetCode 771
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.
题目中文描述
本题大意是，从石头中找珠宝，那么问题来了，什么是石头？ 什么又是珠宝呢？
题目告诉我们，现在你有一串字符，如 aAAbbbb（这串字符就是石头），然后呢，珠宝就是事先定义好的 aA（珠宝），在这里，我们可以看到 ‘石头’ 中有 ‘珠宝’ 为 3 个，输出 3 即可
解题思路
石头由字符串 S 给出，珠宝由字符串 J 给出
- 计算 S 和 J 的各自长度
- 遍历 S 中的字符，和 J 逐一比较即可（反之亦可）
C语言实现
# include <stdio.h>
# include <string.h>
int numJewelsInStones(char *J, char *S)
{
int sum = 0;
int j = 0;
int s = 0;
int j_length = strlen(J);
int s_length = strlen(S);
for (j = 0; j < j_length; j ++ )
{
for ( s = 0; s < s_length; s ++)
{
if (J[j] == S[s])
{
sum++;
}
}
}
return sum;
}
main()
{
int sum = 0;
char j[50];
char s[50];
scanf("%s",&j);
scanf("%s",&s);
sum = numJewelsInStones(j, s);
printf("sum = %d\n", sum);
}
