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1080 · Max Area of Island - LintCode

# Description

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:

input:
[[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]]
output : 6.
Explanation : Note the answer is not 11, because the island must be connected 4-directionally.

Example 2:

input: [[0,0,0,0,0,0,0,0]]
output : 0

# BFS Solution

public class Solution {
    /**
     * @param grid: a 2D array
     * @return: the maximum area of an island in the given 2D array
     */
    public int maxAreaOfIsland(int[][] grid) {
        
        boolean[][] isVisited = new boolean[grid.length][grid[0].length];
        int result = 0;
        for (int i = 0; i < grid.length; i++){
            for (int j = 0; j < grid[0].length; j++){
                if (!isVisited[i][j] && grid[i][j] == 1) {
                    isVisited[i][j] = true;
                    result = Math.max(bfs(grid, isVisited, i, j), result);
                } 
            }
        }
        return result;
    }
    public int bfs(int[][] grid, boolean[][] isVisited, int startx, int endy){
        int[] dx = {0, 0, 1, -1};
        int[] dy = {1, -1, 0, 0};
        Queue<Integer> queue = new LinkedList<>();
        int size = 1;
        queue.offer(startx);
        queue.offer(endy);
        while (!queue.isEmpty()){
            int curx = queue.poll();
            int cury = queue.poll();
            for (int i = 0; i < 4; i++){
              int nextx = curx + dx[i];
              int nexty = cury + dy[i];
              if (nextx >= 0 && nextx < grid.length && 
                  nexty >= 0 && nexty < grid[0].length &&
                  !isVisited[nextx][nexty] && grid[nextx][nexty] == 1) {
                      queue.offer(nextx);
                      queue.offer(nexty);
                      isVisited[nextx][nexty] = true;
                      size++;
                  }
            }
        }
        return size;
    }
}
# DFS Solution
a
public class Solution {
    /**
     * @param grid: a 2D array
     * @return: the maximum area of an island in the given 2D array
     * 通过 dfs 计算每个岛的大小
     */
    public int maxAreaOfIsland(int[][] grid) {
        // Write your code here
        int res = 0;
        for(int i = 0; i < grid.length; i++)
            for(int j = 0; j < grid[0].length; j++)
                if (grid[i][j] == 1) {
                    res = Math.max(res, dfs(grid, i, j));
                }
        return res;
    }
    
    public int dfs(int[][] grid, int i, int j) {
        if( i >= 0 && i < grid.length && j >= 0 && j < grid[0].length && grid[i][j] == 1){
            grid[i][j] = 0;
            return 1 + dfs(grid, i + 1, j) + dfs(grid, i - 1, j) + dfs(grid, i, j - 1) + dfs(grid, i, j + 1);
        }
        return 0;
    }
}