0.FloodFill简介

dfs：深度优先遍历（红色）

bfs：宽度优先遍历

1.图像渲染

算法原理

class Solution {
    int[] dx = { 0, 0, 1, -1 };
    int[] dy = { 1, -1, 0, 0 };

    public int[][] floodFill(int[][] image, int sr, int sc, int color) {
        int prev = image[sr][sc]; // 统计刚开始的颜⾊
        if (prev == color)
            return image; // 处理边界情况
        int m = image.length, n = image[0].length;
        Queue<int[]> q = new LinkedList<>();
        q.add(new int[] { sr, sc });
        while (!q.isEmpty()) {
            int[] t = q.poll();
            int a = t[0], b = t[1];
            image[a][b] = color;
            // 上下左右四个⽅向
            for (int i = 0; i < 4; i++) {
                int x = a + dx[i], y = b + dy[i];
                if (x >= 0 && x < m && y >= 0 && y < n && image[x][y] == prev) {
                    q.add(new int[] { x, y });
                }
            }
        }
        return image;
    }
}

 2.岛屿数量

200. 岛屿数量 - 力扣（LeetCode）

算法原理 

class Solution {
    int[] dx = { 0, 0, -1, 1 };
    int[] dy = { 1, -1, 0, 0 };
    boolean[][] vis = new boolean[301][301];
    int m, n;

    public int numIslands(char[][] grid) {
        m = grid.length;
        n = grid[0].length;
        int ret = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == '1' && !vis[i][j]) {
                    ret++;
                    bfs(grid, i, j);
                }
            }
        }
        return ret;
    }

    public void bfs(char[][] grid, int i, int j) {
        Queue<int[]> q = new LinkedList<>();
        q.add(new int[] { i, j });
        vis[i][j] = true;
        while (!q.isEmpty()) {
            int[] t = q.poll();
            int a = t[0], b = t[1];
            for (int k = 0; k < 4; k++) {
                int x = a + dx[k], y = b + dy[k];
                if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == '1' &&
                        !vis[x][y]) {
                    q.add(new int[] { x, y });
                    vis[x][y] = true;
                }
            }
        }
    }
}

 3.岛屿的最大面积

695. 岛屿的最大面积 - 力扣（LeetCode）

class Solution {
    int[] dx = { 0, 0, 1, -1 };
    int[] dy = { 1, -1, 0, 0 };
    boolean[][] vis = new boolean[51][51];
    int m, n;

    public int maxAreaOfIsland(int[][] grid) {
        m = grid.length;
        n = grid[0].length;
        int ret = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1 && !vis[i][j]) {
                    ret = Math.max(ret, bfs(grid, i, j));
                }
            }
        }
        return ret;
    }

    public int bfs(int[][] grid, int i, int j) {
        int count = 0;
        Queue<int[]> q = new LinkedList<>();
        q.add(new int[] { i, j });
        vis[i][j] = true;
        count++;
        while (!q.isEmpty()) {
            int[] t = q.poll();
            int a = t[0], b = t[1];
            for (int k = 0; k < 4; k++) {
                int x = a + dx[k], y = b + dy[k];
                if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == 1 &&
                        !vis[x][y]) {
                    q.offer(new int[] { x, y });
                    vis[x][y] = true;
                    count++;
                }
            }
        }
        return count;
    }
}

 4.被围绕的区域

130. 被围绕的区域 - 力扣（LeetCode）

算法原理

class Solution {
    int n, m;

    public void solve(char[][] board) {
        n = board.length;
        if (n == 0) {
            return;
        }
        m = board[0].length;
        for (int i = 0; i < n; i++) {
            dfs(board, i, 0);
            dfs(board, i, m - 1);
        }
        for (int i = 1; i < m - 1; i++) {
            dfs(board, 0, i);
            dfs(board, n - 1, i);
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (board[i][j] == 'A') {
                    board[i][j] = 'O';
                } else if (board[i][j] == 'O') {
                    board[i][j] = 'X';
                }
            }
        }
    }

    public void dfs(char[][] board, int x, int y) {
        if (x < 0 || x >= n || y < 0 || y >= m || board[x][y] != 'O') {
            return;
        }
        board[x][y] = 'A';
        dfs(board, x + 1, y);
        dfs(board, x - 1, y);
        dfs(board, x, y + 1);
        dfs(board, x, y - 1);
    }
}