LeetCode #771 JewelsAndStones. Algorithm,알고리즘,LeetCode,Codefights,CodeSignal,코드파이트,코드시그널,예제,문제해결능력,example,c++,java,재귀,recursive,datastructure,techinterview,coding,코딩인터뷰,기술면접,연결리스트
어려운건 어려운대로 잘 안풀려서 스트레스
쉬운건 쉬운대로 너무 쉽게 풀려서 스트레스
Difficult problem is so hard to solve, it makes me mad
Easy problem is too easy to handle it, it also makes me mad
God...d
Runtime: 4 ms, faster than 86.23% of C++ online submissions for Jewels and Stones.
Memory Usage: 8.3 MB, less than 79.28% of C++ online submissions for Jewels and Stones.
LeetCode #771
Q.
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".
보석인 돌을 나타내는 문자열 J 를 입력받고, 본인이 갖고 있는 돌을 나타내는 문자열 S 를 입력받는다. S 의 각 문자는 본인이 가진 돌의 타입이다. 갖고 있는 돌 중에 몇개나 보석인지 알아보고 싶다. 문자열 J 는 다 구별되어있고, J 와 S 는 문자들이다. 문자들은 각각 A, a 대소문자 구별된다.
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.
Process
// Process
//1. Input String S and J
//2. Make jewel alphabet plate using String J
//3. Iterate from begin to the end (String S)
// 3.1. Check if my stone is jewel or not
//4. Return count
// 처리과정
//1. 문자열 S, J 입력받는다.
//2. 보석 알파벳 다이나믹플레이트 만든다 (문자열 J)
//3. 시작부터 끝까지 반복한다 (문자열 S)
// 3.1. 내 돌이 보석인지 아닌지 확인한다
//4. 보석카운트 반환한다.
Code.. lemme see example code!!!
코드.. 예제코드를 보자!!!
class Solution {
public:
int numJewelsInStones(string J, string S) {
int letterPlate[58] = { 0 };
int jewelCount = 0;
for (int i = 0; i < J.size(); ++i) {
++letterPlate[J[i]-65];
}
for (int i = 0; i < S.size(); ++i) {
if (letterPlate[S[i]-65] > 0)
++jewelCount;
}
return jewelCount;
}
};
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