LeetCode #383 RansomNote. Algorithm,알고리즘,LeetCode,Codefights,CodeSignal,코드파이트,코드시그널,예제,문제해결능력,example,c++,java,재귀,recursive,datastructure,techinterview,coding,코딩인터뷰,기술면접,연결리스트
Performance 99%
Memory both 99%%
LeetCode #383
Q.
Given an arbitrary ransom note string and another string containing letters from all the magazines, write a function that will return true if the ransom note can be constructed from the magazines ; otherwise, it will return false.
Each letter in the magazine string can only be used once in your ransom note.
주어진 임의의 랜섬노트 문자열과, 모든 매거진 글자들을 포함하는 문자열이 있다. 매거진의 모든 글자들을 사용해서 랜섬노트 문자열을 만들수 있는지 확인하는 함수를 만들어라. 되면 true, 안되면 false
매거진의 문자열은 랜섬노트를 만들때 딱 한번만 사용할 수 있다.
Note:
You may assume that both strings contain only lowercase letters.
둘다 소문자들로만 구성된 문자열이다.
canConstruct("a", "b") -> false
canConstruct("aa", "ab") -> false
canConstruct("aa", "aab") -> true
Process
// Process
//1. Input ransomNote, magazine
//2. Iterate Magazine
// 2.1 Make dynamicLower
//3. Iterate ransomNote
// 3.1. If dynamicLower[i] is bigger than 0,
// 3.1.1. Subtract 1
// 3.2. If not -> it's false
//4. Return result
// 처리과정
//1. 랜섬노트, 매거진 문자열을 입력받는다.
//2. 매거진 문자열 반복한다.
// 2.1. 소문자동적배열 만들어둔다.
//3. 랜섬노트 반복한다.
// 3.1. 소문자동적배열의 해당 위치가 0보다 크면
// 3.1.1. 1뺀다.
// 3.2. 작으면
// 3.2.1. false
//4. Return result
Code.. lemme see example code!!!
코드.. 예제코드를 보자!!!
class Solution {
public:
bool canConstruct(string ransomNote, string magazine) {
bool result = true;
int dynamicLower[26] = { 0, };
for (int i = 0; i < magazine.size(); ++i) {
++dynamicLower[magazine[i] - 97];
}
for (int i = 0; result && i < ransomNote.size(); ++i) {
if (dynamicLower[ransomNote[i] - 97] > 0) {
--dynamicLower[ransomNote[i] - 97];
}
else {
result = false;
}
}
return result;
}
};
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