JAVA刷题记录: 专题十五 BFS解决FloodFill算法

733. 图像渲染 - 力扣（LeetCode）

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];
        if(color == prev) return image;
        Queue q = new LinkedList<>();
        q.add(new int[]{sr, sc});
        int m = image.length, n = image[0].length;
        while(!q.isEmpty()) {
            int[] tmp = q.poll();
            image[tmp[0]][tmp[1]] = color;
            for(int i = 0; i < 4; i++) {
                int x = tmp[0] + dx[i], y = tmp[1] + dy[i];
                if(x >= 0 && x < m && y >= 0 && y < n && image[x][y] == prev) {
                    q.add(new int[]{x, y});
                }
            }
        }
        return image;
    }
}

200. 岛屿数量 - 力扣（LeetCode）

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

    public int numIslands(char[][] grid) {
        m = grid.length; n = grid[0].length;
        vis = new boolean[m][n];
        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]) {
                    bfs(grid, i, j);
                    ret++;
                }
            } 
        }
        return ret;
    }

    public void bfs(char[][] grid, int i, int j) {
        Queue 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;
                }
            }
        }
    }
}

695. 岛屿的最大面积 - 力扣（LeetCode）

class Solution {
    int m, n;
    boolean[][] vis;
    int[] dx = {0, 0, -1, 1};
    int[] dy = {1, -1, 0, 0};
    public int maxAreaOfIsland(int[][] grid) {
        int ret = 0;
        m = grid.length; n = grid[0].length;
        vis = new boolean[m][n];
        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 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.add(new int[]{x, y});
                    vis[x][y] = true;
                    count++;
                }
            } 
        }
        return count;
    }
}

130. 被围绕的区域 - 力扣（LeetCode）

class Solution {
    int m, n;
    int[] dx = {0, 0, -1, 1};
    int[] dy = {-1, 1, 0, 0};
    public void solve(char[][] board) {
        m = board.length; n = board[0].length;
        for(int i = 0; i < m; i++) {
            if(board[i][0] == 'O') bfs(board, i, 0);
            if(board[i][n - 1] == 'O') bfs(board, i, n - 1);
        }
        for(int i = 0; i < n; i++) {
            if(board[0][i] == 'O') bfs(board, 0, i);
            if(board[m - 1][i] == 'O') bfs(board, m - 1, i);
        }

        for(int i = 0; i < m; i++) {
            for(int j = 0; j < n; j++) {
                if(board[i][j] == 'O') board[i][j] = 'X';
                else if(board[i][j] == '.') board[i][j] = 'O';
            }
        }
    }

    public void bfs(char[][] board, int i, int j) {
        Queue q = new LinkedList<>();
        q.add(new int[]{i, j});
        board[i][j] = '.';
        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 && board[x][y] == 'O') {
                    q.add(new int[]{x, y});
                    board[x][y] = '.';
                }
            } 
        }
    }
}