[Algorithm_Java] 광물 캐기 (Success)

프로그래머스

 

☕Language: Java

 

🚨 곡갱이 순서 부여 로직 생성 실패

import java.util.Arrays;
 
class Solution {
    static int[] mineralsToInt;
    static int[] sumWeight;
    public int solution(int[] picks, String[] minerals) {
        // minerals: 5
        // diamond count sort
        // picks 0 1 2
        int answer = 0;
        
        calcWeightMinerals(minerals);
        calcSumWeight(mineralsToInt);
        createOrderPickUse(picks);
        Arrays.sort(sumWeight);
                
        return answer;
    }
    public static int[] calcWeightMinerals(String[] minerals){
        mineralsToInt = new int[minerals.length];
        for(int i=0; i<minerals.length; i++){
            switch(minerals[i]){
                case "diamond": mineralsToInt[i]=3; break;
                case "iron": mineralsToInt[i]=2; break;
                case "stone": mineralsToInt[i]=1; break; 
            }
        }
        return mineralsToInt;
    }
    public static int[] calcSumWeight(int[] mineralsToInt){
        sumWeight = new int[(mineralsToInt.length/5)+1];
        int sumSubWeight = 0;
        int idx=0;
        for(int i=0; i<mineralsToInt.length; i++){
            sumSubWeight += mineralsToInt[i];
            if(i%5==4 || i==(mineralsToInt.length-1)){
                sumWeight[idx++] = sumSubWeight;
                sumSubWeight = 0;
            }
        }
        return sumWeight;
    }
}
 

 

😮 찾아본 풀이

⭐이중 배열과 포인터 활용

 

import java.util.*;
 
class Solution {
    public int solution(int[] picks, String[] minerals) {
        int answer = 0;
        int countPicks = picks[0] + picks[1] + picks[2];
        int[][] mineralFatigueTable = new int[minerals.length / 5 + 1][3];
        
        for(int i = 0; i < minerals.length && countPicks > 0; i++) {
            switch(minerals[i])
            {
                case "diamond":
                    mineralFatigueTable[i / 5][0] += 1;
                    mineralFatigueTable[i / 5][1] += 5;
                    mineralFatigueTable[i / 5][2] += 25;
                    break;
                case "iron":
                    mineralFatigueTable[i / 5][0] += 1;
                    mineralFatigueTable[i / 5][1] += 1;
                    mineralFatigueTable[i / 5][2] += 5;
                    break;
                case "stone":
                    mineralFatigueTable[i / 5][0] += 1;
                    mineralFatigueTable[i / 5][1] += 1;
                    mineralFatigueTable[i / 5][2] += 1;
            }
            
            if(i % 5 == 4) countPicks--;            
            
        }
        
        Arrays.sort(mineralFatigueTable, new Comparator<int[]>() {
            @Override
            public int compare(int[] o1, int[] o2) {
                if(o1[2] < o2[2]) 
                    return 1;
                else 
                    return -1;
            }
        });
        
        for(int i = 0, pickIndx = 0; i < mineralFatigueTable.length; i++) {
            while(pickIndx < 3 && picks[pickIndx] == 0) pickIndx++;
            if(pickIndx == 3) break;
            picks[pickIndx]--;
            answer += mineralFatigueTable[i][pickIndx];
        }
        
        return answer;
    }
}
 

 

1. mineralFatigueTable

    - [광물을 5개씩 묶었을 때 몇 번째 묶음인지,곡괭이 별 피로도]

2. Arrays.sort(mineralFatigueTable)

    - compare

        - 피로도 계산이 가장 큰 돌을 사용했을 때 기준으로 5개 광물 캐기 피로도 묶음을 내림차순 정렬

    - pickIdx로 mineralFatigueTable에서 피로도 선택

 

 

 

🔗 소스 코드
GitHub

 

📚 참고 자료

프로그래머스 광물 캐기 java