code-squad · honux77 · Oct 15, 2020 · Feb 3, 2020 · Feb 3, 2020 · Feb 4, 2020
diff --git a/200211/102092/groupby1.sql b/200211/102092/groupby1.sql
@@ -0,0 +1,5 @@
+/*
+https://programmers.co.kr/learn/courses/30/lessons/59040
+*/
+
+SELECT ANIMAL_TYPE,COUNT(ANIMAL_TYPE) FROM ANIMAL_INS GROUP BY ANIMAL_TYPE ORDER BY ANIMAL_TYPE;
diff --git a/200211/102092/workoutClothes.java b/200211/102092/workoutClothes.java
@@ -0,0 +1,85 @@
+import org.junit.jupiter.api.BeforeEach;
+import org.junit.jupiter.api.Test;
+
+import static org.junit.jupiter.api.Assertions.assertEquals;
+
+/*
+https://programmers.co.kr/learn/courses/30/lessons/42862
+
+1. Greedy 문제
+2. 다중 for문을 돌려야하나?
+3. 정확히 어떤 컨셉의 문제인지 빠르게 캐치가 안됨.
+4.
+ */
+
+class Solution {
+
+    public int solution(int n, int[] lost, int[] reserve) {
+        int result = n;
+        result -= lost.length;
+
+        for (int i = 0; i < lost.length; i++) {
+            for (int j = 0; j < reserve.length; j++) {
+                // 여분을 가져왔는데, 잃어버렸을 경우, 그러면 빌려줄 수 없으니까.
+                if (lost[i] == reserve[j]) {
+                    result++;
+                    lost[i] = -1;
+                    reserve[j] = -1;
+                    break;
+                }
+            }
+
+        }
+
+        for (int i = 0; i < lost.length; i++) {
+            for (int j = 0; j < reserve.length; j++) {
+                if (lost[i] == reserve[j] - 1 || lost[i] == reserve[j] + 1) {
+                    result++;
+                    reserve[j] = -1;
+                    break;
+                }
+            }
+
+        }
+        return result;
+    }
+}
+
+public class workoutClothes {
+
+    Solution solution;
+
+    @BeforeEach
+    public void 객체생성() {
+        solution = new Solution();
+    }
+
+    @Test
+    public void 테스트1() {
+        int n = 5;
+        int[] lost = new int[] { 2, 4 };
+        int[] reserve = new int[] { 1, 3, 5 };
+        int result = 5;
+        assertEquals(solution.solution(n, lost, reserve), result);
+    }
+
+    @Test
+    public void 테스트2() {
+        int n = 5;
+        int[] lost = new int[] { 2, 4 };
+        int[] reserve = new int[] { 3 };
+        int result = 4;
+
+        assertEquals(solution.solution(n, lost, reserve), result);
+    }
+
+    @Test
+    public void 테스트3() {
+        int n = 3;
+        int[] lost = new int[] { 3 };
+        int[] reserve = new int[] { 1 };
+        int result = 2;
+
+        assertEquals(solution.solution(n, lost, reserve), result);
+    }
+}
diff --git a/200212/102092/MoreSpicy.java b/200212/102092/MoreSpicy.java
@@ -0,0 +1,31 @@
+import java.util.PriorityQueue;
+
+//https://programmers.co.kr/learn/courses/30/lessons/42626
+
+/*
+1. 스코빌 지수가, 숫자대로 정렬되어야 한다는 점에서 우선순위 큐를 이용하여 풀었음.
+
+*/
+
+class Solution {
+    public int solution(int[] scoville, int K) {
+        PriorityQueue<Integer> pq = new PriorityQueue<>();
+        int answer = 0;
+        for(int input : scoville){
+            pq.add(input);
+        }
+
+        while(pq.peek() < K){
+            if(pq.size() == 1){
+                return -1;
+            }
+
+            int lowest = pq.poll();
+            int slowest = pq.poll();
+            int mix = lowest + (slowest * 2);
+            pq.add(mix);
+            answer++;
+        }
+        return answer;
+    }
+}
diff --git a/200212/102092/groupby2.sql b/200212/102092/groupby2.sql
@@ -0,0 +1,11 @@
+/*
+https://programmers.co.kr/learn/courses/30/lessons/59041
+
+SELECT NAME, COUNT(*) FROM ANIMAL_INS GROUP BY NAME  HAVING COUNT(*) > 1;
+null인 경우를 신경 쓰지 않아서 실패함..
+
+SELECT NAME, COUNT(*) FROM ANIMAL_INS WHERE NAME IS NOT NULL GROUP BY NAME  HAVING COUNT(NAME) > 1;
+
+*/
+
+SELECT NAME, COUNT(NAME) FROM ANIMAL_INS GROUP BY NAME  HAVING COUNT(NAME) > 1;
diff --git a/200213/102092/RomanToInteger.java b/200213/102092/RomanToInteger.java
@@ -0,0 +1,40 @@
+/*
+https://leetcode.com/problems/roman-to-integer/
+*/
+
+class Solution {
+
+    public int romanToInt(String s) {
+        int sum = 0;
+        if (s.contains("IV"))
+            sum -= 2;
+        if (s.contains("IX"))
+            sum -= 2;
+        if (s.contains("XL"))
+            sum -= 20;
+        if (s.contains("XC"))
+            sum -= 20;
+        if (s.contains("CD"))
+            sum -= 200;
+        if (s.contains("CM"))
+            sum -= 200;
+
+        for (int i = 0; i < s.length(); i++) {
+            if (s.charAt(i) == 'I')
+                sum += 1;
+            if (s.charAt(i) == 'V')
+                sum += 5;
+            if (s.charAt(i) == 'X')
+                sum += 10;
+            if (s.charAt(i) == 'L')
+                sum += 50;
+            if (s.charAt(i) == 'C')
+                sum += 100;
+            if (s.charAt(i) == 'D')
+                sum += 500;
+            if (s.charAt(i) == 'M')
+                sum += 1000;
+        }
+        return sum;
+    }
+}
diff --git a/200213/102092/groupby3.sql b/200213/102092/groupby3.sql
@@ -0,0 +1,5 @@
+/*
+https://programmers.co.kr/learn/courses/30/lessons/59412
+*/
+
+SELECT HOUR(DATETIME), COUNT(DATETIME) FROM ANIMAL_OUTS WHERE HOUR(DATETIME)BETWEEN '09:00:00' AND '19:00:00' GROUP BY HOUR(DATETIME);
diff --git a/200214/102092/IsNull.sql b/200214/102092/IsNull.sql
@@ -0,0 +1,5 @@
+/*
+https://programmers.co.kr/learn/courses/30/lessons/59039
+*/
+
+SELECT ANIMAL_ID FROM ANIMAL_INS WHERE NAME IS NULL; 
diff --git a/200214/102092/ValidParentheses.java b/200214/102092/ValidParentheses.java
@@ -0,0 +1,37 @@
+/*
+https://leetcode.com/problems/valid-parentheses/
+*/
+
+import java.util.HashMap;
+import java.util.Stack;
+
+public class ValidParentheses {
+
+    Solution solution;
+
+    class Solution {
+
+        public boolean isValid(String s) {
+            HashMap<Character, Character> hashMap = new HashMap<>();
+            hashMap.put('(', ')');
+            hashMap.put('[', ']');
+            hashMap.put('{', '}');
+
+            Stack<Character> stack = new Stack();
+
+            for (int i = 0; i < s.length(); i++) {
+                char target = s.charAt(i);
+                if (target == '(' || target == '[' || target == '{') {
+                    stack.add(target);
+                } else {
+                    if (stack.isEmpty() || hashMap.get(stack.peek()) != target) {
+                        return false;
+                    }
+                    stack.pop();
+                }
+
+            }
+            return stack.isEmpty();
+        }
+    }
+}
diff --git a/200217/102092/IsNull2.sql b/200217/102092/IsNull2.sql
@@ -0,0 +1,5 @@
+/*
+https://programmers.co.kr/learn/courses/30/lessons/59407
+*/
+
+SELECT ANIMAL_ID FROM ANIMAL_INS WHERE NAME IS NOT NULL;
diff --git a/200217/102092/MergeTwoSorted.java b/200217/102092/MergeTwoSorted.java
@@ -0,0 +1,66 @@
+/*
+ * https://leetcode.com/problems/merge-two-sorted-lists/submissions/
+ * 
+ * 
+ * public ListNode mergeTwoLists(ListNode l1, ListNode l2){
+		if(l1 == null) return l2;
+		if(l2 == null) return l1;
+		if(l1.val < l2.val){
+			l1.next = mergeTwoLists(l1.next, l2);
+			return l1;
+		} else{
+			l2.next = mergeTwoLists(l1, l2.next);
+			return l2;
+		}
+}
+- 회귀를 이용해서 푸는 방법도 있구나.
+ */
+
+class Solution {
+
+    public class ListNode {
+        int val;
+        ListNode next;
+
+        ListNode(int x) {
+            val = x;
+        }
+    }
+
+    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
+        ListNode answer, target;
+
+        if (l1 == null && l2 == null) {
+            return null;
+        }
+        if (l1 == null) {
+            return l2;
+        }
+        if (l2 == null) {
+            return l1;
+        }
+
+        answer = new ListNode(0);
+        target = answer;
+
+        while (l1 != null && l2 != null) {
+            if (l1.val >= l2.val) {
+                target.next = l2;
+                l2 = l2.next;
+            } else {
+                target.next = l1;
+                l1 = l1.next;
+            }
+            target = target.next;
+        }
+
+        if (l1 != null) {
+            target.next = l1;
+        }
+        if (l2 != null) {
+            target.next = l2;
+        }
+        return answer.next;
+
+    }
+}
diff --git a/200218/102092/IsNull3.sql b/200218/102092/IsNull3.sql
@@ -0,0 +1,5 @@
+/*
+https://programmers.co.kr/learn/courses/30/lessons/59410
+*/
+
+SELECT ANIMAL_TYPE, IFNULL(NAME,'No name'), SEX_UPON_INTAKE FROM ANIMAL_INS;
diff --git a/200218/102092/MergeSortedArray.java b/200218/102092/MergeSortedArray.java
@@ -0,0 +1,36 @@
+/*
+https://leetcode.com/problems/merge-sorted-array/submissions/
+*/
+
+import java.util.Arrays;
+
+class Solution {
+    public void merge(int[] nums1, int m, int[] nums2, int n) {
+        int index = 0;
+        for (int i = 0; i < nums1.length; i++) {
+            if (nums1[i] == 0 && index < n) {
+                nums1[i] = nums2[index];
+                index++;
+            }
+        }
+        Arrays.sort(nums1);
+    }
+}
+
+/*
+ * - 출제자가 의도한 해답은 아니었는 듯. 
+ * - 처음 생각은 빈곳에 num2 배열 값을 넣고, Arrays.sort를 돌리자
+ * 
+ */
+
+// public class Solution2 {
+//     public void merge(int A[], int m, int B[], int n) {
+//         int i = m - 1, j = n - 1, k = m + n - 1;
+//         while (i >= 0 && j >= 0) {
+//             A[k--] = A[i] > B[j] ? A[i--] : B[j--];
+//         }
+//         while (j >= 0) {
+//             A[k--] = B[j--];
+//         }
+//     }
+// }
diff --git a/200219/102092/Join1.sql b/200219/102092/Join1.sql
@@ -0,0 +1,7 @@
+/*
+https://programmers.co.kr/learn/courses/30/lessons/59042
+*/
+
+SELECT OUTS.ANIMAL_ID , OUTS.NAME FROM ANIMAL_INS AS INS
+RIGHT JOIN ANIMAL_OUTS AS OUTS ON INS.ANIMAL_ID = OUTS.ANIMAL_ID
+WHERE INS.ANIMAL_ID IS NULL;