week3/Vreyz11

src/week3/거스름돈/Vryez11/Solution.java

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+package week3.거스름돈.Vryez11;
+
+public class Solution {
+
+    /**
+     *
+     * [프로그래머스] 거스름돈
+     *
+     * 문제 난이도: Lv. 3
+     * 문제 링크: https://school.programmers.co.kr/learn/courses/30/lessons/12907
+     * 풀이 시간: 20분
+     * 풀이 근거: 타일 개수 더하기랑 비슷함, 1원으로 0원에서 1원, 1원에서 1원 더하면 2원 ... DP
+     */
+
+    final static int MOD = 1_000_000_007;
+
+    public int solution(int n, int[] money) {
+
+        int[] dp = new int[n + 1];
+        dp[0] = 1;
+
+        for (int coin : money) {
+            for (int price = coin; price <= n; price++) {
+
+                dp[price] = (dp[price] + dp[price - coin]) % MOD;
+            }
+        }
+
+        return dp[n];
+    }
+}

src/week3/뉴스클러스터링/Vryez11/Solution.java

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+package week3.뉴스클러스터링.Vryez11;
+
+import java.util.HashMap;
+import java.util.Map;
+
+public class Solution {
+
+    /**
+     *
+     * [프로그래머스] 뉴스 클러스터링
+     *
+     * 문제 난이도: Lv. 2
+     * 문제 링크: https://school.programmers.co.kr/learn/courses/30/lessons/17677
+     * 풀이 시간: 30분
+     * 풀이 근거: Map을 통해 2자 씩 put / 이를 통해 합집합, 교집합 구하기
+     */
+
+    public int solution(String str1, String str2) {
+
+        Map<String, Integer> str1Map = new HashMap<>();
+        Map<String, Integer> str2Map = new HashMap<>();
+
+        str1 = str1.toLowerCase();
+        str2 = str2.toLowerCase();
+
+        for (int i = 0; i < str1.length() - 1; i++) {
+
+            char char1 = str1.charAt(i);
+            char char2 = str1.charAt(i + 1);
+
+            if (!isAlphabetChar(char1, char2)) {
+                continue;
+            }
+
+            String word = (char1 + "" + char2);
+
+            str1Map.put(word, str1Map.getOrDefault(word, 0) + 1);
+        }
+        for (int i = 0; i < str2.length() - 1; i++) {
+
+            char char1 = str2.charAt(i);
+            char char2 = str2.charAt(i + 1);
+
+            if (!isAlphabetChar(char1, char2)) {
+                continue;
+            }
+
+            String word = (char1 + "" + char2);
+
+            str2Map.put(word, str2Map.getOrDefault(word, 0) + 1);
+        }
+
+        int union = 0;
+        int intersection = 0;
+
+        for (String key : str1Map.keySet()) {
+
+            intersection += str1Map.get(key);
+
+            if (str2Map.containsKey(key)) {
+                union += Math.min(str1Map.get(key), str2Map.get(key));
+            }
+        }
+
+        for (String key : str2Map.keySet()) {
+
+            intersection += str2Map.get(key);
+        }
+
+        System.out.println("union = " + union);
+        System.out.println("intersection = " + intersection);
+
+        intersection -= union;
+
+        if (union == 0 && intersection == 0) {
+            return 65_536;
+        }
+
+        return (int) (((double) union / intersection) * 65_536);
+    }
+
+    private boolean isAlphabetChar(char char1, char char2) {
+        return char1 <= 'z' && char1 >= 'a' && 'a' <= char2 && char2 <= 'z';
+    }
+}

src/week3/도넛과막대그래프/Solution.java

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+package week3.도넛과막대그래프;
+
+
+class Solution {
+
+    /**
+     *
+     * [프로그래머스] 도넛과 막대 그래프
+     *
+     * 문제 난이도: Lv 2
+     * 문제 링크: https://school.programmers.co.kr/learn/courses/30/lessons/258711
+     * 풀이 시간: 못 품..
+     * 풀이 근거: 문제를 잘 읽자.., in, out 생각을 못했습니다...
+     */
+
+    public int[] solution(int[][] edges) {
+
+        int max = 0;
+
+        for (int[] edge : edges) {
+            max = Math.max(max, Math.max(edge[0], edge[1]));
+        }
+
+        int[] in = new int[max + 1];
+        int[] out = new int[max + 1];
+
+        for (int[] edge : edges) {
+            out[edge[0]]++;
+            in[edge[1]]++;
+        }
+
+        int create = 0;
+        int stick = 0;
+        int eight = 0;
+
+        for (int i = 1; i <= max; i++) {
+
+            if (in[i] == 0 && out[i] >= 2) {
+                create = i;
+            }
+
+            if (in[i] >= 1 && out[i] == 0) {
+                stick++;
+            }
+
+            if (in[i] >= 2 && out[i] == 2) {
+                eight++;
+            }
+        }
+
+        int donut = out[create] - stick - eight;
+
+        return new int[]{create, donut, stick, eight};
+    }
+}

src/week3/여행결로/Vryez11/Solution.java

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+package week3.여행결로.Vryez11;
+
+import java.util.*;
+
+public class Solution {
+
+    /**
+     *
+     * [프로그래머스] 여행경로
+     *
+     * 문제 난이도: LV. 3
+     * 문제 링크: https://school.programmers.co.kr/learn/courses/30/lessons/43164
+     * 풀이 시간: 30분
+     * 풀이 근거: 티켓을 먼저 정렬시켜서, 한 공항에서 다른 여러 공항이 있을 때, 사전 순으로 가도록 함. 그 다음 dfs로 해결
+     */
+
+    private Map<String, Integer> portNumMap;
+    private List<Integer>[] lists;
+
+    public String[] solution(String[][] tickets) {
+
+        Arrays.sort(tickets, (o1, o2) -> {
+            if (o1[0].equals(o2[0])) {
+                return o1[1].compareTo(o2[1]);
+            }
+            return o1[0].compareTo(o2[0]);
+        });
+
+        portNumMap = new HashMap<>();
+        int idx = 0;
+
+        for (String[] ticket : tickets) {
+
+            if (!portNumMap.containsKey(ticket[0])) {
+                portNumMap.put(ticket[0], idx++);
+            }
+
+            if (!portNumMap.containsKey(ticket[1])) {
+                portNumMap.put(ticket[1], idx++);
+            }
+        }
+
+        lists = new ArrayList[idx];
+
+        for (int i = 0; i < idx; i++) {
+            lists[i] = new ArrayList<>();
+        }
+
+        for (int i = 0; i < tickets.length; i++) {
+            int from = portNumMap.get(tickets[i][0]);
+            lists[from].add(i);
+        }
+
+        boolean[] visited = new boolean[tickets.length];
+        List<String> answer = new ArrayList<>();
+
+        dfs("ICN", 0, tickets, visited, answer);
+
+        return answer.toArray(new String[0]);
+    }
+
+    private boolean dfs(String now,
+                        int depth,
+                        String[][] tickets,
+                        boolean[] visited,
+                        List<String> answer) {
+
+        answer.add(now);
+
+        if (depth == tickets.length) {
+            return true;
+        }
+
+        Integer nowNum = portNumMap.get(now);
+
+        if (nowNum == null) {
+            answer.remove(answer.size() - 1);
+            return false;
+        }
+
+        for (int ticketIdx : lists[nowNum]) {
+
+            if (visited[ticketIdx]) {
+                continue;
+            }
+
+            visited[ticketIdx] = true;
+
+            if (dfs(tickets[ticketIdx][1],
+                    depth + 1,
+                    tickets,
+                    visited,
+                    answer)) {
+                return true;
+            }
+
+            visited[ticketIdx] = false;
+        }
+
+        answer.remove(answer.size() - 1);
+
+        return false;
+    }
+}

src/week3/점프와순간이동/Vryez11/Solution.java

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+package week3.점프와순간이동.Vryez11;
+
+public class Solution {
+
+    /**
+     *
+     * [프로그래머스] 점프와 순간 이동
+     *
+     * 문제 난이도: Lv2
+     * 문제 링크: https://school.programmers.co.kr/learn/courses/30/lessons/12980
+     * 풀이 시간: 20분
+     * 풀이 근거: 2로 나눠지면 순간 이동 / 아니면 한 칸 앞으로 가기
+     */
+
+    public int solution(int n) {
+
+        int battery = 0;
+
+        while (n > 0) {
+            if (n % 2 == 0) {
+                n /= 2;
+            } else {
+                n -= 1;
+                battery += 1;
+            }
+        }
+
+        return battery;
+    }
+}