Merge pull request #2263 from se6816/main

[se6816] WEEK 09 solutions

Author: se6816

linked-list-cycle/se6816.java

@@ -0,0 +1,40 @@
+/**
+    길이가 최대 10000이므로 10001까지 경우의 수를 돌아 확인하는 방법
+ */
+public class Solution {
+    public boolean hasCycle(ListNode head) {
+        int max = 10001;
+        int idx = 0;
+        boolean result = false;
+        while(head != null) {
+            idx++;
+            head = head.next;
+            if(idx > max) {
+                result = true;
+                break;
+            }
+        }
+
+        return result;
+    }
+}
+/**
+    2개의 포인터를 이용하여, 만나는지 확인하는 방법
+ */
+public class Solution2 {
+    public boolean hasCycle(ListNode head) {
+        ListNode first = head;
+        ListNode second = head;
+        boolean result = false;
+        while(second != null && second.next != null) {
+            first = first.next;
+            second = second.next.next;
+            if(first == second) {
+                result = true;
+                break;
+            }
+        }
+
+        return result;
+    }
+}

number-of-islands/se6816.java

@@ -0,0 +1,52 @@
+/**
+    방문하지 않는 지점을 찾고, BFS로 인근 지점을 방문처리하며, 방문하지 않는 지점 개수를 카운틷하는 방식
+ */
+class Solution {
+    int M;
+    int N;
+    boolean[][] visited;
+    int[] moveX = {-1, 0, 0, 1};
+    int[] moveY = {0, -1, 1, 0};
+    public int numIslands(char[][] grid) {
+        int result = 0;
+        M = grid.length;
+        N = grid[0].length;
+        visited= new boolean[M][N];
+        for(int i = 0; i < M; i++) {
+            for(int j = 0; j < N; j++) {
+                if(!visited[i][j] && grid[i][j] == '1') {
+                    visited[i][j] = true;
+                    result++;
+                    paint(i, j, visited, grid);
+                }
+            }
+        }
+        return result;
+        
+    }
+    public void paint(int x, int y, boolean[][] visited, char[][] grid) {
+        for(int i = 0; i < 4; i++) {
+            int tempX = x + moveX[i];
+            int tempY = y + moveY[i];
+
+            if(isOutOfIndex(tempX, tempY)) {
+                continue;
+            }
+
+            if(visited[tempX][tempY]) {
+                continue;
+            }
+
+            if(grid[tempX][tempY] == '0') {
+                continue;
+            }
+
+            visited[tempX][tempY] = true;
+
+            paint(tempX,tempY, visited, grid);
+        }
+    }
+    public boolean isOutOfIndex(int x, int y) {
+        return x < 0 || x >= M || y < 0 || y >=N;
+    }
+}

pacific-atlantic-water-flow/se6816.java

@@ -0,0 +1,73 @@
+/**
+    BFS를 통해, 방문 지점이 겹치는 부분을 찾는 방식
+ */
+class Solution {
+    public static int[] moveX = {0, 1, 0, -1};
+    public static int[] moveY = {1, 0, -1, 0};
+    public boolean[][] POvisited;
+    public boolean[][] AOvisited;
+    public int N;
+    public int M;
+    public List<List<Integer>> pacificAtlantic(int[][] heights) {
+        List<List<Integer>> result = new ArrayList<>();
+        N = heights.length;
+        M = heights[0].length;
+        POvisited = new boolean[N][M];
+        AOvisited = new boolean[N][M];
+        Queue<int[]> que = new LinkedList<>();
+        Queue<int[]> que2 = new LinkedList<>();
+
+        for(int i = 0; i < M; i++) {
+            POvisited[0][i] = true;
+            que.add(new int[]{0, i});
+            AOvisited[N-1][i] = true;
+            que2.add(new int[]{N-1, i});
+        }
+
+        for(int i = 0; i < N; i++) {
+            POvisited[i][0] = true;
+            que.add(new int[]{i, 0});
+            AOvisited[i][M-1] = true;
+            que2.add(new int[]{i, M-1});
+        }
+        bfs(que, POvisited, heights);
+        bfs(que2, AOvisited, heights);
+
+        for(int i = 0; i < N; i++) {
+            for(int j = 0; j < M; j++) {
+                if(POvisited[i][j] && AOvisited[i][j]) {
+                    result.add(new ArrayList<>(List.of(i, j)));
+                }
+            }
+        }
+
+        return result;
+    }
+
+    public void bfs(Queue<int[]> que, boolean[][] visited, int[][] heights) {
+        while(!que.isEmpty()) {
+            int[] node = que.poll();
+            for(int i = 0; i < 4; i++) {
+                int tempX = node[0] + moveX[i];
+                int tempY = node[1] + moveY[i];
+                if(isOutOfIndex(tempX, tempY)) {
+                    continue;
+                }
+
+                if(visited[tempX][tempY]) {
+                    continue;
+                }
+                if(heights[tempX][tempY] < heights[node[0]][node[1]]) {
+                    continue;
+                }
+
+                visited[tempX][tempY] = true;
+                que.add(new int[]{tempX, tempY});
+            }
+        }
+    }
+
+    public boolean isOutOfIndex(int x, int y){
+        return x < 0 || x >= N || y < 0 || y >= M;
+    }
+}

unique-paths/se6816.java

@@ -0,0 +1,48 @@
+class Solution {
+    public int[] moveX = {0, 1};
+    public int[] moveY = {1, 0};
+    int N;
+    int M;
+    public int uniquePaths(int m, int n) {
+        int[][] dp = new int[m][n];
+        boolean[][] visited = new boolean[m][n];
+        int curX = 0;
+        int curY = 0;
+        N = n;
+        M = m;
+        dp[curX][curY] = 1;
+        visited[curX][curY] = true;
+        Queue<int[]> que = new ArrayDeque<>();
+        que.offer(new int[]{0,0});
+
+        while(!que.isEmpty()) {
+            int[] data = que.poll();
+
+            for(int i = 0; i < 2; i++) {
+                int tempX = data[0] + moveX[i];
+                int tempY = data[1] + moveY[i];
+
+                if(isOutOfIndex(tempX, tempY)) {
+                    continue;
+                }
+
+                dp[tempX][tempY] += dp[data[0]][data[1]];
+
+
+                if(visited[tempX][tempY]) {
+                    continue;
+                }
+                visited[tempX][tempY] = true;
+                que.add(new int[]{tempX, tempY});
+
+            }
+        }
+
+        return dp[M-1][N-1];
+
+    }
+
+    public boolean isOutOfIndex(int x, int y) {
+        return x < 0 || x >= M || y < 0 || y >=N;
+    }
+}