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:
11110
11010
11000
00000
Answer: 1
Example 2:
11000
11000
00100
00011
Answer: 3
dfs
public class Solution { public int numIslands(char[][] grid) { 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; } public void dfs(char[][] grid, int x, int y){ if( x < 0 || x > grid.length - 1 || y< 0 || y > grid[0].length - 1 || grid[x][y] == '0') return; grid[x][y] = '0'; dfs(grid, x+1, y); dfs(grid, x-1, y); dfs(grid, x, y-1); dfs(grid, x, y+1); }
bfs
public class Solution { public int numIslands(char[][] grid) { int res = 0; Queue<int[]> queue = new LinkedList<>(); for(int i = 0; i < grid.length; i++){ for(int j = 0 ; j < grid[0].length; j++){ if(grid[i][j] == '1'){ res++; queue.add(new int[]{i,j}); while(!queue.isEmpty()){ int point[] = queue.poll(); int row = point[0]; int col = point[1]; if(grid[row][col] == '0') continue; grid[row][col] = '0'; if(row > 0) queue.add(new int[]{row-1,col}); if(col > 0 ) queue.add(new int[]{row, col-1}); if(row < grid.length -1) queue.add(new int[]{row+1, col}); if(col < grid[0].length -1) queue.add(new int[]{row, col+1}); } } } } return res; } }