用于查找岛蟒的 DFS

Question

I have been trying this approach for a little while and keep getting 0 for the count of islands. I am not quite sure what I am doing wrong here. As I have started learning python more and more so I may be missing something that is easy to spot.

So I want to find all the islands with a 1 in them and if they are connected (adjacent cells). So I want to return the count of the number of islands that are found.

Here is what I have so far.

def valid_direction(A, r, c):
    row = len(A)
    col = len(A[0])
    if r < 0 or c < 0 or r >= row or c >= col:
        return False
    else:
        return True

def dfs(A, r, c):
    A[r][c] = '1'
    # Up down left and right
    directions = [(0,1), 
                  (0,-1), 
                  (-1,0),
                  (1,0)] 
    for d in directions:
        nr = r +d[0]
        nc = c + d[1]
        if valid_direction(A, nr, nc) and A[nr][nc] == '1':
            dfs(A, nr, nc)


def solution(A):
    if not A:
        return -1

    row = len(A)
    col = len(A[0])
    results = 0
    for i in range(row):
        for j in range(col):
            if A[i][j] == '1':
                dfs(A, i, j)
                results +=1
    return results

And here are the two arrays I am working with

A1 = [[1,1,1,1,0],
      [1,1,0,1,0],
      [1,1,0,0,0,],
      [0,0,0,0,0]]

A2 = [[1,1,0,0,0],
      [1,1,0,0,0],
      [0,0,1,0,0],
      [0,0,0,1,1]]

Answer 1

You made a simple mistake. Your arrays are arrays of integers, not chars.
A[r][c] = '1' and A[nr][nc] == '1' and if A[i][j] == '1'
should be
A[r][c] = 1 and A[nr][nc] == 1 and if A[i][j] == 1
I'm not sure it will solve the question correctly after this, but this is the current error.

---

Your current code just counts the number of 1s.
You should check if the 1 is visited ex)by changing it to 2.
Also you don't have to initiate directions every time dfs is called you should just take it outside the function.

directions = [(0,1), 
              (0,-1), 
              (-1,0),
              (1,0)]

def valid_direction(A, r, c):
    row = len(A)
    col = len(A[0])
    if r < 0 or c < 0 or r >= row or c >= col:
        return False
    else:
        return True

def dfs(A, r, c):
    A[r][c] = 2
    # Up down left and right 
    for d in directions:
        nr = r +d[0]
        nc = c + d[1]
        if valid_direction(A, nr, nc) and A[nr][nc] == 1:
            dfs(A, nr, nc)


def solution(A):
    if not A:
        return -1

    row = len(A)
    col = len(A[0])
    results = 0
    for i in range(row):
        for j in range(col):
            if A[i][j] == 1:
                dfs(A, i, j)
                results +=1
    return results

I'm not completely sure what you are trying to solve, but I think the correct code is this.