老卫带你学---leetcode刷题(200. 岛屿数量)

200. 岛屿数量

问题：

给你一个由 ‘1’（陆地）和 ‘0’（水）组成的的二维网格，请你计算网格中岛屿的数量。

岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。

此外，你可以假设该网格的四条边均被水包围。

示例 1：

输入：grid = [
  ["1","1","1","1","0"],
  ["1","1","0","1","0"],
  ["1","1","0","0","0"],
  ["0","0","0","0","0"]
]
输出：1

示例 2：

输入：grid = [
  ["1","1","0","0","0"],
  ["1","1","0","0","0"],
  ["0","0","1","0","0"],
  ["0","0","0","1","1"]
]
输出：3

提示：

m == grid.length
n == grid[i].length
1 <= m, n <= 300
grid[i][j] 的值为 '0' 或 '1'

解决：

dfs搜索

逐一遍历每个元素，然后进行dfs搜索

dfs搜索就是（上下左右四个方向，如果超过边界或者为0停止搜索）

但是在搜索的过程中需要注意将已经搜过的置为0，避免重复搜索

class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        count =0
        def dfs(i,j):
            if not 0<=i<len(grid) or not 0<=j<len(grid[0]) or grid[i][j]=="0":
                return
            grid[i][j]="0"
            dfs(i,j-1)
            dfs(i,j+1)
            dfs(i-1,j)
            dfs(i+1,j)
        
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if grid[i][j]=="1":
                    dfs(i,j)
                    count+=1
        return count