200. Number of Islands / 695. Max Area of Island

200. Number of Islands

Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Example 1:

Input:
11110
11010
11000
00000

Output: 1

Example 2:

Input:
11000
11000
00100
00011

Output: 3

My Solutions:

在dfs中遇到一个为1的岛，标记岛为2，防止重复计算

class Solution {
    public int numIslands(char[][] grid) {
        if (grid == null || grid.length == 0) return 0;
        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') {
                    dfs(grid, i, j);
                    res++;
                }
            }
        }
        return res;
    }
    
    private void dfs(char[][] grid, int i, int j) {
        if (i < 0 || i >= grid.length || j < 0 || j >= grid[i].length) return;
        if (grid[i][j] == '1') {
            grid[i][j] = '2';
            dfs(grid, i + 1, j);
            dfs(grid, i - 1, j);
            dfs(grid, i, j + 1);
            dfs(grid, i, j - 1);
        }
    }
}

695. Max Area of Island

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.

Example 2:

[[0,0,0,0,0,0,0,0]]

Given the above grid, return 0.

Note: The length of each dimension in the given grid does not exceed 50.

My Solutions:

和上题思想相同，每一次dfs完一个1的所有连接岛屿后，更新result的大小

class Solution {
    public int maxAreaOfIsland(int[][] grid) {
        if (grid == null || grid.length == 0) return 0;
        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) {
                    int max = dfs(grid, i, j);
                    if (max > res) res = max;
                }
            }
        }
        return res;
    }
    
    private int dfs(int[][] grid, int i, int j) {
        if (i < 0 || i >= grid.length || j < 0 || j >= grid[i].length || grid[i][j] != 1) return 0;
        grid[i][j] = 2;
        return 1 + dfs(grid, i + 1, j) 
            + dfs(grid, i - 1, j)
            + dfs(grid, i, j + 1)
            + dfs(grid, i, j - 1);
    }
}