https://leetcode.com/problems/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:
11110
11010
11000
00000
Output: 1
- code
class Solution {
void dfs(char[][] grid, int r, int c) {
int nr = grid.length;
int nc = grid[0].length;
if (r < 0 || c < 0 || r >= nr || c >= nc || grid[r][c] == '0') {
return;
}
grid[r][c] = '0';
dfs(grid, r - 1, c);
dfs(grid, r + 1, c);
dfs(grid, r, c - 1);
dfs(grid, r, c + 1);
}
public int numIslands(char[][] grid) {
if (grid == null || grid.length == 0) {
return 0;
}
int nr = grid.length;
int nc = grid[0].length;
int num_islands = 0;
for (int r = 0; r < nr; ++r) {
for (int c = 0; c < nc; ++c) {
if (grid[r][c] == '1') {
++num_islands;
dfs(grid, r, c);
}
}
}
return num_islands;
}
}
- code
class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
self.m = len(grid)
self.n = len(grid[0])
count = 0
for row in range(self.m):
for col in range(self.n):
if grid[row][col] == "1":
count += 1
self.dfs(grid, row, col)
return count
def dfs(self, grid, x, y):
grid[x][y] = "0"
directions = [[1,0],[-1,0],[0,1],[0,-1]]
for dx, dy in directions:
if 0 <= x + dx < self.m and 0 <= y + dy < self.n and grid[x+dx][y+dy] == "1":
self.dfs(grid, x + dx, y + dy)