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Max Area of Island

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Question

Given a non-empty 2D array grid of 0’s and 1’s, an island is a group of 1’s (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.

Find the maximum area of an island in the given 2D array. (If there is no island, the maximum area is 0.)

Example 1:
[[0,0,1,0,0,0,0,1,0,0,0,0,0],
[0,0,0,0,0,0,0,1,1,1,0,0,0],
[0,1,1,0,1,0,0,0,0,0,0,0,0],
[0,1,0,0,1,1,0,0,1,0,1,0,0],
[0,1,0,0,1,1,0,0,1,1,1,0,0],
[0,0,0,0,0,0,0,0,0,0,1,0,0],
[0,0,0,0,0,0,0,1,1,1,0,0,0],
[0,0,0,0,0,0,0,1,1,0,0,0,0]]
Given the above grid, return 6. Note the answer is not 11, because the island must be connected 4-directionally

Solution 1

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/*
dfs
*/    
int row;
int col;
int result;
int count;
public int maxAreaOfIsland(int[][] grid) {
    if (grid.length == 0 || grid[0].length == 0) {
        return 0;
    }
    
    row = grid.length;
    col = grid[0].length;
    boolean[][] visited = new boolean[row][col];
    result = 0;
    for (int i = 0; i < row; i++) {
        for (int j = 0; j < col; j++) {
            if (grid[i][j] == 1 && !visited[i][j]) {
                count = 0;
                dfs(i, j, grid, visited);
                result = Math.max(result, count);
            }
        }
    }        
    return result;
}

public void dfs(int i, int j, int[][] grid, boolean[][] visited) {
    if (i < 0 || j < 0 || i >= row || j >= col || visited[i][j] || grid[i][j] == 0) {
        return;
    }
    visited[i][j] = true;
    count++;
    dfs(i+1, j, grid, visited);
    dfs(i-1, j, grid, visited);
    dfs(i, j+1, grid, visited);
    dfs(i, j-1, grid, visited);
}

Solution 2

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/*
BFS
*/
class Node {
    int x;
    int y;
    Node(int x, int y) {
        this.x = x;
        this.y = y;
    }
}
int num;
int row;
int col;
public int maxAreaOfIsland(int[][] grid) {
    if (grid.length == 0 || grid[0].length == 0) {
        return 0;
    }
    row = grid.length;
    col = grid[0].length;
    Queue<Node> queue = new LinkedList<>();
    boolean[][] visited = new boolean[row][col];
    
    for (int i = 0; i < row; i++) {
        for (int j = 0; j < col; j++) {
            if (grid[i][j] == 1 && !visited[i][j]) {
                queue.add(new Node(i,j));
                visited[i][j] = true;
                BFS(queue, visited, grid);
            }
        }
    }
    return num;
}

public void BFS(Queue<Node> queue, boolean[][] visited, int[][] grid) {
    int[] dirX = {0,0,-1,1};
    int[] dirY = {1,-1,0,0};
    
    int count = 1;
    while (!queue.isEmpty()) {
        Node cur = queue.poll();
        for (int i = 0; i < 4; i++) {
            int nextX = cur.x+dirX[i];
            int nextY = cur.y+dirY[i];
            if (isValid(nextX, nextY, grid, visited)) {
                visited[nextX][nextY] = true;
                queue.add(new Node(nextX, nextY));
                count++;
            }
        }
    }
    num = Math.max(num, count);
}

public boolean isValid(int x, int y, int[][] grid, boolean[][] visited) {
    if (x >= 0 && y >= 0 && x < row && y < col && !visited[x][y] && grid[x][y] == 1) {
        return true;
    }
    return false;
}