add: week 7 solutions

Author: Yongseok-Choi-dev

longest-substring-without-repeating-characters/YoungSeok-Choi.java

@@ -0,0 +1,40 @@
+import java.util.HashMap;
+import java.util.Map;
+
+class Solution {
+    public int lengthOfLongestSubstring(String s) {
+        int mx = 0;
+        int autoIncrement = 0;
+        Map<Character, Integer> map = new HashMap<>();
+        Map<Integer, Character> reverseMap = new HashMap<>();
+
+        for (char c : s.toCharArray()) {
+            ++autoIncrement;
+            if (map.containsKey(c)) {
+                mx = Math.max(mx, map.size());
+
+                int start = map.get(c);
+                // NOTE: 중복 문자를 만나는 경우, 중복된 문자 이전에 Map에 들어온 원소는 모두 제거, (e.g. "sadvdf" 라는 입력이
+                // 들어왔을 때 sad까지만 제거가 되고 v는 남아있어야 함.)
+                for (int i = start; i >= 1; i--) {
+                    if (reverseMap.containsKey(i)) {
+                        char target = reverseMap.get(i);
+                        reverseMap.remove(i);
+                        map.remove(target);
+
+                    } else {
+                        break;
+                    }
+                }
+
+                map.put(c, autoIncrement);
+                reverseMap.put(autoIncrement, c);
+            } else {
+                map.put(c, autoIncrement);
+                reverseMap.put(autoIncrement, c);
+            }
+        }
+
+        return Math.max(mx, map.size());
+    }
+}

number-of-islands/YoungSeok-Choi.java

@@ -0,0 +1,78 @@
+import java.util.LinkedList;
+import java.util.Queue;
+
+class Solution {
+
+    public int[] dx = { 1, 0, -1, 0 };
+    public int[] dy = { 0, 1, 0, -1 };
+    public int cnt = 0;
+    public int w = 0;
+    public int h = 0;
+    public boolean[][] visit;
+    public Queue<Node> q = new LinkedList<>();
+
+    public int numIslands(char[][] grid) {
+        w = grid.length;
+        h = grid[0].length;
+        visit = new boolean[w][h];
+
+        for (int i = 0; i < w; i++) {
+            for (int j = 0; j < h; j++) {
+                if (grid[i][j] == '1') {
+                    cnt++;
+                    // dfs(grid, i, j);
+                    bfs(grid, i, j);
+                }
+            }
+        }
+
+        return cnt;
+    }
+
+    public void dfs(char[][] grid, int x, int y) {
+        if (x < 0 || x >= w || y < 0 || y >= h || grid[x][y] == '0' || visit[x][y]) {
+            return;
+        }
+
+        visit[x][y] = true;
+        grid[x][y] = '0';
+
+        for (int i = 0; i < 4; i++) {
+            int nx = x + dx[i];
+            int ny = y + dy[i];
+
+            dfs(grid, nx, ny);
+        }
+    }
+
+    public void bfs(char[][] grid, int x, int y) {
+
+        q.add(new Node(x, y));
+
+        while (!q.isEmpty()) {
+            Node p = q.poll();
+
+            for (int i = 0; i < 4; i++) {
+                int nx = p.x + dx[i];
+                int ny = p.y + dy[i];
+
+                if (x < 0 || x >= w || y < 0 || y >= h || grid[x][y] == '0' || visit[x][y]) {
+                    continue;
+                }
+
+                grid[nx][ny] = '0';
+                q.add(new Node(nx, ny));
+            }
+        }
+    }
+}
+
+class Node {
+    int x;
+    int y;
+
+    public Node(int x, int y) {
+        this.x = x;
+        this.y = y;
+    }
+}

reverse-linked-list/YoungSeok-Choi.java

@@ -0,0 +1,82 @@
+import java.util.HashMap;
+import java.util.Map;
+
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ * int val;
+ * ListNode next;
+ * ListNode() {}
+ * ListNode(int val) { this.val = val; }
+ * ListNode(int val, ListNode next) { this.val = val; this.next = next; }
+ * }
+ */
+class Solution {
+    public ListNode reverseList(ListNode head) {
+        int idx = 0;
+        Map<Integer, Integer> idxMap = new HashMap<>();
+
+        if (head == null || head.next == null) {
+            return null;
+        }
+
+        while (true) {
+            idxMap.put(idx++, head.val);
+
+            if (head.next == null) {
+                break;
+            }
+
+            head = head.next;
+        }
+
+        ListNode resHead = new ListNode();
+        ListNode cur = resHead;
+
+        for (int i = idx - 1; i >= 0; i--) {
+            cur.val = idxMap.get(i);
+
+            if (i != 0) {
+                cur.next = new ListNode();
+                cur = cur.next;
+            }
+        }
+
+        return resHead;
+    }
+
+}
+
+// O(n) + 변수 하나로 직관적으로 문제를 해결할 수 있다..
+class AnotherSolution {
+    public ListNode reverstList(ListNode head) {
+        ListNode prev = null;
+        ListNode cur = head;
+
+        while (cur.next != null) {
+            ListNode next = cur.next;
+            cur.next = prev;
+            prev = cur;
+            cur = next;
+        }
+
+        return prev;
+    }
+}
+
+class ListNode {
+    int val;
+    ListNode next;
+
+    ListNode() {
+    }
+
+    ListNode(int val) {
+        this.val = val;
+    }
+
+    ListNode(int val, ListNode next) {
+        this.val = val;
+        this.next = next;
+    }
+}

set-matrix-zeroes/YoungSeok-Choi.java

@@ -0,0 +1,54 @@
+import java.util.ArrayList;
+import java.util.List;
+
+class Solution {
+
+  public List<Node> point = new ArrayList<>();
+
+  public void setZeroes(int[][] matrix) {
+
+    for (int i = 0; i < matrix.length; i++) {
+      for (int j = 0; j < matrix[0].length; j++) {
+        if (matrix[i][j] == 0) {
+          point.add(new Node(i, j));
+        }
+      }
+    }
+
+    for (Node n : point) {
+      int x = n.x;
+      int y = n.y;
+
+      int nx = n.x;
+      while (nx > 0) {
+        matrix[--nx][y] = 0;
+      }
+
+      int ny = n.y;
+      while (ny > 0) {
+        matrix[x][--ny] = 0;
+      }
+
+      int nnx = n.x;
+      while (nnx < matrix.length - 1) {
+        matrix[++nnx][y] = 0;
+      }
+
+      int nny = n.y;
+      while (nny < matrix[0].length - 1) {
+        matrix[x][++nny] = 0;
+      }
+    }
+  }
+
+}
+
+class Node {
+  int x;
+  int y;
+
+  public Node(int x, int y) {
+    this.x = x;
+    this.y = y;
+  }
+}

unique-paths/YoungSeok-Choi.java

@@ -0,0 +1,31 @@
+// NOTE: 2 * X 부터 모든 경우를 세어보니, 3 * 3의 격자가 나왔을 때 2 * 3, 3 * 2 경우를 더한 값이 나오는 것을 알 수 있었음
+// +)  3 * 4 격자에 대한 답은 3 * 3 격자의 답 + 2 * 4 격자의 답을 더하면 만들어지는 것을 알수 있었고, 아래와 같은 점화식을 만들 수 있었음
+// memo[m][n] = memo[m][n - 1] + memo[m - 1][n]
+class Solution {
+  public int uniquePaths(int m, int n) {
+    int[][] memo = new int[101][101];
+
+    if (m == 1 || n == 1) {
+      return 1;
+    }
+
+    memo[2][2] = 2;
+    for (int i = 2; i < 3; i++) {
+      for (int j = 3; j < 101; j++) {
+        memo[i][j] = j;
+        memo[j][i] = j;
+      }
+    }
+
+    for (int i = 3; i < 101; i++) {
+      memo[i][i] = memo[i][i - 1] + memo[i - 1][i];
+      for (int j = i + 1; j < 101; j++) {
+        int val = memo[i][j - 1] + memo[i - 1][j];
+        memo[i][j] = val;
+        memo[j][i] = val;
+      }
+    }
+
+    return memo[m][n];
+  }
+}