771 Jewels and Stones 宝石与石头
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:
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.
题目描述:
给定字符串J 代表石头中宝石的类型，和字符串 S代表你拥有的石头。 S 中每个字符代表了一种你拥有的石头的类型，你想知道你拥有的石头中有多少是宝石。
J 中的字母不重复，J 和 S中的所有字符都是字母。字母区分大小写，因此"a"和"A"是不同类型的石头。
示例 :
示例 1:
输入: J = "aA", S = "aAAbbbb"
输出: 3
示例 2:
输入: J = "z", S = "ZZ"
输出: 0
注意：
S 和 J 最多含有50个字母。
J 中的字符不重复。
思路:
- 用数组记录下宝石的种类, 遍历 S即可
- 正则表达式
时间复杂度O(n ^ 2), 空间复杂度O(1)
代码:
C++:
class Solution
{
public:
int numJewelsInStones(string J, string S)
{
int count[128] = {0}, result = 0;
for (char c : J) ++count[c];
for (char c : S) if (count[c]) ++result;
return result;
}
};
Java:
class Solution {
public int numJewelsInStones(String J, String S) {
return S.replaceAll("[^" + J + "]", "").length();
}
}
Python:
class Solution:
def numJewelsInStones(self, J: str, S: str) -> int:
return len([i for i in S if i in J])
