Programers 대충 만든 자판/ C++, JAVA

https://school.programmers.co.kr/learn/courses/30/lessons/160586?language=cpp 

프로그래머스

 

아이디어

알파벳에 따른 최소 입력 키 횟수를 저장해두고 targets에 있는 알파벳에 따라 입력하는 키의 횟수를 더하면 된다.

HashMap 사용

 

 C++

#include <string>
#include <vector>
#include <unordered_map>
using namespace std;

vector<int> solution(vector<string> keymap, vector<string> targets) {
    vector<int> answer;
    unordered_map<char,int> key;
    for(auto iter : keymap){
        for(int i=0;i<iter.size();i++){
            if(key.find(iter[i]) == key.end()){
                key[iter[i]] = i+1;
                continue;
            }
            if(key[iter[i]] > i+1)
                key[iter[i]] = i+1;
        }
    }
    for(auto iter : targets){
        int keyCnt = 0;
        for(int i=0;i<iter.size();i++){
            if(key.find(iter[i]) == key.end()){
                keyCnt = -1;
                break;
            }
            keyCnt += key[iter[i]];                
        }
        answer.push_back(keyCnt);
    }
    
    return answer;
}

 

JAVA

import java.util.*;
class Solution {
    public int[] solution(String[] keymap, String[] targets) {
        int[] answer = new int[targets.length];
        HashMap<Character,Integer> map = new HashMap<>();

        for (String key : keymap) {
            for (int i = 0; i < key.length(); i++) {
                char ch = key.charAt(i);
                if (!map.containsKey(ch)){
                    map.put(ch,i+1);
                    continue;
                }
                if(map.get(ch)>i){
                    map.replace(ch,i+1);
                }
                
            }
        }

        for(int i=0; i< targets.length;i++){
            int sum = 0;
            for(int j=0; j<targets[i].length();j++){

                char ch = targets[i].charAt(j);

                if(!map.containsKey(ch)){
                    sum = -1;
                    break;
                }
                sum+=map.get(ch);
            }
            answer[i] = sum;
        }

        return answer;
    }
}