题解 | #岛屿数量#

深度优先搜索

#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 判断岛屿数量
# @param grid char字符型二维数组 
# @return int整型
#
class Solution:
    def solve(self , grid: List[List[str]]) -> int:
        # write code here
        if not grid:
            return 0
        cnt = 0
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if grid[i][j] == '1':
                    cnt += 1
                    self.dfs(grid, i, j)
        return cnt
                    
                    
        
    # 深度优先搜索
    def dfs(self, grid: List[List[str]], i: int, j: int):
        grid[i][j] = '0'
        n = len(grid)
        m = len(grid[0])
        if i - 1 >= 0 and grid[i - 1][j] == '1':
            self.dfs(grid, i - 1, j)
        if i + 1 < n and grid[i + 1][j] == '1':
            self.dfs(grid, i + 1, j)
        if j - 1 >= 0 and grid[i][j - 1] == '1':
            self.dfs(grid, i, j - 1)
        if j + 1 < m and grid[i][j + 1] == '1':        
            self.dfs(grid, i, j + 1)