[leetcode] 200. Number of Islands

题目

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.

An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Example 1:

Input: grid = [
  ["1","1","1","1","0"],
  ["1","1","0","1","0"],
  ["1","1","0","0","0"],
  ["0","0","0","0","0"]
]
Output: 1

Example 2:

Input: grid = [
  ["1","1","0","0","0"],
  ["1","1","0","0","0"],
  ["0","0","1","0","0"],
  ["0","0","0","1","1"]
]
Output: 3

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 300
• grid[i][j] is '0' or '1'.

思路

dfs，遍历二维数组，当遇到“1”时，使用dfs覆盖周围所有的“1”，并增加计数，直至遍历结束。

代码

python版本：

class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        def fill(x, y):
            if x < 0 or x >= len(grid) or y < 0 or y >= len(grid[0]) or grid[x][y] == '0':
                return
            grid[x][y] = '0'
            for l, r in [[0, 1], [0, -1], [1, 0], [-1, 0]]:
                fill(x+l, y+r)
        cnt = 0
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if grid[i][j] == '1':
                    cnt += 1
                    fill(i, j)
        return cnt

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