代码随想录算法训练营第五十二天 | 101. 孤岛的总面积 102.沉没孤岛 103.水流问题 104.建造最大岛屿

101. 孤岛的总面积

题目链接：101. 孤岛的总面积

import java.util.*;

public class Main {
    private static int count = 0;
    private static final int[][] dir = {{0, 1}, {1, 0}, {-1, 0}, {0, -1}}; // 四个方向

    private static void bfs(int[][] grid, int x, int y) {
        Queue<int[]> que = new LinkedList<>();
        que.add(new int[]{x, y});
        grid[x][y] = 0; // 只要加入队列，立刻标记
        count++;
        while (!que.isEmpty()) {
            int[] cur = que.poll();
            int curx = cur[0];
            int cury = cur[1];
            for (int i = 0; i < 4; i++) {
                int nextx = curx + dir[i][0];
                int nexty = cury + dir[i][1];
                if (nextx < 0 || nextx >= grid.length || nexty < 0 || nexty >= grid[0].length) continue; // 越界了，直接跳过
                if (grid[nextx][nexty] == 1) {
                    que.add(new int[]{nextx, nexty});
                    count++;
                    grid[nextx][nexty] = 0; // 只要加入队列立刻标记
                }
            }
        }
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int m = scanner.nextInt();
        int[][] grid = new int[n][m];
        
        // 读取网格
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                grid[i][j] = scanner.nextInt();
            }
        }
        
        // 从左侧边，和右侧边向中间遍历
        for (int i = 0; i < n; i++) {
            if (grid[i][0] == 1) bfs(grid, i, 0);
            if (grid[i][m - 1] == 1) bfs(grid, i, m - 1);
        }
        
        // 从上边和下边向中间遍历
        for (int j = 0; j < m; j++) {
            if (grid[0][j] == 1) bfs(grid, 0, j);
            if (grid[n - 1][j] == 1) bfs(grid, n - 1, j);
        }
        
        count = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j] == 1) bfs(grid, i, j);
            }
        }

        System.out.println(count);
    }
}

102.沉没孤岛

题目链接：102. 沉没孤岛

import java.util.Scanner;

public class Main {
    static int[][] dir = { {-1, 0}, {0, -1}, {1, 0}, {0, 1} }; // 保存四个方向

    public static void dfs(int[][] grid, int x, int y) {
        grid[x][y] = 2;
        for (int[] d : dir) {
            int nextX = x + d[0];
            int nextY = y + d[1];
            // 超过边界
            if (nextX < 0 || nextX >= grid.length || nextY < 0 || nextY >= grid[0].length) continue;
            // 不符合条件，不继续遍历
            if (grid[nextX][nextY] == 0 || grid[nextX][nextY] == 2) continue;
            dfs(grid, nextX, nextY);
        }
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int m = scanner.nextInt();
        int[][] grid = new int[n][m];

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                grid[i][j] = scanner.nextInt();
            }
        }

        // 步骤一：
        // 从左侧边，和右侧边 向中间遍历
        for (int i = 0; i < n; i++) {
            if (grid[i][0] == 1) dfs(grid, i, 0);
            if (grid[i][m - 1] == 1) dfs(grid, i, m - 1);
        }

        // 从上边和下边 向中间遍历
        for (int j = 0; j < m; j++) {
            if (grid[0][j] == 1) dfs(grid, 0, j);
            if (grid[n - 1][j] == 1) dfs(grid, n - 1, j);
        }

        // 步骤二、步骤三
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j] == 1) grid[i][j] = 0;
                if (grid[i][j] == 2) grid[i][j] = 1;
            }
        }

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                System.out.print(grid[i][j] + " ");
            }
            System.out.println();
        }
        
        scanner.close();
    }
}

103.水流问题

题目链接：103. 水流问题

import java.util.*;
public class Main {

    // 采用 DFS 进行搜索
    public static void dfs(int[][] heights, int x, int y, boolean[][] visited, int preH) {
        // 遇到边界或者访问过的点，直接返回
        if (x < 0 || x >= heights.length || y < 0 || y >= heights[0].length || visited[x][y]) return;
        // 不满足水流入条件的直接返回
        if (heights[x][y] < preH) return;
        // 满足条件，设置为true，表示可以从边界到达此位置
        visited[x][y] = true;

        // 向下一层继续搜索
        dfs(heights, x + 1, y, visited, heights[x][y]);
        dfs(heights, x - 1, y, visited, heights[x][y]);
        dfs(heights, x, y + 1, visited, heights[x][y]);
        dfs(heights, x, y - 1, visited, heights[x][y]);
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int m = sc.nextInt();
        int n = sc.nextInt();

        int[][] heights = new int[m][n];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                heights[i][j] = sc.nextInt();
            }
        }

        // 初始化两个二位boolean数组，代表两个边界
        boolean[][] pacific = new boolean[m][n];
        boolean[][] atlantic = new boolean[m][n];

        // 从左右边界出发进行DFS
        for (int i = 0; i < m; i++) {
            dfs(heights, i, 0, pacific, Integer.MIN_VALUE);
            dfs(heights, i, n - 1, atlantic, Integer.MIN_VALUE);
        }

        // 从上下边界出发进行DFS
        for (int j = 0; j < n; j++) {
            dfs(heights, 0, j, pacific, Integer.MIN_VALUE);
            dfs(heights, m - 1, j, atlantic, Integer.MIN_VALUE);
        }

        // 当两个边界二维数组在某个位置都为true时，符合题目要求
        List<List<Integer>> res = new ArrayList<>();
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (pacific[i][j] && atlantic[i][j]) {
                    res.add(Arrays.asList(i, j));
                }
            }
        }

        // 打印结果
        for (List<Integer> list : res) {
            for (int k = 0; k < list.size(); k++) {
                if (k == 0) {
                    System.out.print(list.get(k) + " ");
                } else {
                    System.out.print(list.get(k));
                }
            }
            System.out.println();
        }
    }
}

104.建造最大岛屿

题目链接：104. 建造最大岛屿

import java.util.*;
public class Main {
    // 该方法采用 DFS
    // 定义全局变量
    // 记录每次每个岛屿的面积
    static int count;
    // 对每个岛屿进行标记
    static int mark;
    // 定义二维数组表示四个方位
    static int[][] dirs = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};

    // DFS 进行搜索，将每个岛屿标记为不同的数字
    public static void dfs(int[][] grid, int x, int y, boolean[][] visited) {
        // 当遇到边界，直接return
        if (x < 0 || x >= grid.length || y < 0 || y >= grid[0].length) return;
        // 遇到已经访问过的或者遇到海水，直接返回
        if (visited[x][y] || grid[x][y] == 0) return;

        visited[x][y] = true;
        count++;
        grid[x][y] = mark;

        // 继续向下层搜索
        dfs(grid, x, y + 1, visited);
        dfs(grid, x, y - 1, visited);
        dfs(grid, x + 1, y, visited);
        dfs(grid, x - 1, y, visited);
    }

    public static void main (String[] args) {
        // 接收输入
        Scanner sc = new Scanner(System.in);
        int m = sc.nextInt();
        int n = sc.nextInt();

        int[][] grid = new int[m][n];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                grid[i][j] = sc.nextInt();
            }
        }

        // 初始化mark变量，从2开始（区别于0水，1岛屿）
        mark = 2;

        // 定义二位boolean数组记录该位置是否被访问
        boolean[][] visited = new boolean[m][n];

        // 定义一个HashMap，记录某片岛屿的标记号和面积
        HashMap<Integer, Integer> getSize = new HashMap<>();

        // 定义一个HashSet，用来判断某一位置水四周是否存在不同标记编号的岛屿
        HashSet<Integer> set = new HashSet<>();

        // 定义一个boolean变量，看看DFS之后，是否全是岛屿
        boolean isAllIsland = true;

        // 遍历二维数组进行DFS搜索，标记每片岛屿的编号，记录对应的面积
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 0) isAllIsland = false;
                if (grid[i][j] == 1) {
                    count = 0;
                    dfs(grid, i, j, visited);
                    getSize.put(mark, count);
                    mark++;
                }
            }
        }

        int result = 0;
        if (isAllIsland) result =  m * n;

        // 对标记完的grid继续遍历，判断每个水位置四周是否有岛屿，并记录下四周不同相邻岛屿面积之和
        // 每次计算完一个水位置周围可能存在的岛屿面积之和，更新下result变量
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 0) {
                    set.clear();
                    // 当前水位置变更为岛屿，所以初始化为1
                    int curSize = 1;

                    for (int[] dir : dirs) {
                        int curRow = i + dir[0];
                        int curCol = j + dir[1];

                        if (curRow < 0 || curRow >= m || curCol < 0 || curCol >= n) continue;
                        int curMark = grid[curRow][curCol];
                        // 如果当前相邻的岛屿已经遍历过或者HashMap中不存在这个编号，继续搜索
                        if (set.contains(curMark) || !getSize.containsKey(curMark)) continue;
                        set.add(curMark);
                        curSize += getSize.get(curMark);
                    }

                    result = Math.max(result, curSize);
                }
            }
        }

        // 打印结果
        System.out.println(result);
    }
}