200.岛屿数量.md

200.岛屿数量

https://leetcode-cn.com/problems/number-of-islands/solution/200dao-yu-shu-liang-ju-zhen-sou-suo-top3-ww71/

难度：中等

题目

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

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

分析

推荐下矩阵搜索的经典三连发：

• 79.单词搜索
• 130.被围绕的区域
• 200.岛屿数量

针对这道题，我们只需要对矩阵进行依次遍历，如果当前grid[x][y] == "1",则启动DFS模式。 找到与之相连的所有1，将其置为0。搜索结束后，我们找到了一个岛屿，岛屿数量+=1。 如此循环，最终返回岛屿数量即可。

解题

class Solution:
    def numIslands(self, grid):
        row, col, ret = len(grid), len(grid[0]), 0

        def dfs(x, y):
            grid[x][y] = '0'
            for c in [[0, 1], [0, -1], [1, 0], [-1, 0]]:
                nx, ny = x + c[0], y + c[1]
                if 0 <= nx < row and 0 <= ny < col and grid[nx][ny] == '1':
                    dfs(nx, ny)

        for i in range(row):
            for j in range(col):
                if grid[i][j] == '1':
                    dfs(i, j)
                    ret += 1
        return ret

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