计算二维矩阵中带变化的岛的数量

def numIslands(grid):
    if not grid:
        return 0
    
    def dfs(i, j):
        if i < 0 or i >= len(grid) or j < 0 or j >= len(grid[0]) or grid[i][j] != '1':
            return
        
        grid[i][j] = '0'
        dfs(i+1, j)
        dfs(i-1, j)
        dfs(i, j+1)
        dfs(i, j-1)
    
    count = 0
    for i in range(len(grid)):
        for j in range(len(grid[0])):
            if grid[i][j] == '1':
                count += 1
                dfs(i, j)
    
    return count

# 示例输入
grid = [
    ['1', '1', '0', '0', '0'],
    ['1', '1', '0', '0', '0'],
    ['0', '0', '1', '0', '0'],
    ['0', '0', '0', '1', '1']
]

# 调用函数计算岛的数量
island_count = numIslands(grid)
print("带变化的岛的数量为:", island_count)