Update README.md

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@ -212,7 +212,7 @@ def iterative_dfs(graph, v, discovered):
---
### matrix bfs
### matrix bfs and dfs
<br>
@ -223,6 +223,7 @@ def iterative_dfs(graph, v, discovered):
* fill empty room with the distance to its nearest gate. if it is impossible to reach a gate, it should be filled with `INF`.
<br>
```python
@ -261,3 +262,78 @@ def matrix_bfs(rooms) -> None:
rooms[r][c] = rooms[row][col] + 1
queue.append((r, c))
```
<br>
* given an `m x n` 2D binary grid grid which represents a map of '1's (land) and '0's (water), return the number of islands.
<br>
```python
def num_island_dfs(grid) -> int:
LAND = '1'
answer = 0
def dfs(row, col):
if row < 0 or row >= len(grid) or col < 0 or col >= len(grid[0]) or grid[row][col] != LAND:
return
grid[row][col] = 'x'
dfs(row + 1, col)
dfs(row - 1, col)
dfs(row, col - 1)
dfs(row, col + 1)
for i in range(len(grid)):
for j in range(len(grid[0])):
if grid[i][j] == LAND:
answer += 1
dfs(i, j)
return answer
```
<br>
* and a solution for the problem above, using bfs:
<br>
```python
def num_island_bfs(grid) -> int:
LAND = '1'
answer = 0
queue = collections.deque()
def bfs(row, col, queue):
delta = [(1, 0), (0, 1), (-1, 0), (0, -1)]
while queue:
x, y = queue.popleft()
for dx, dy in delta:
px, py = x + dx, y + dy
if px < 0 or px >= len(grid) or py < 0 or py >= len(grid[0]) or grid[px][py] != LAND:
continue
grid[px][py] = 'x'
queue.append((px, py))
for i in range(len(grid)):
for j in range(len(grid[0])):
if grid[i][j] == LAND:
answer += 1
queue.append((i, j))
bfs(i, j, queue)
return answer
```