Leetcode200.岛屿数量

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：

1
2
3
4
5
6
7
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：

1
2
3
4
5
6
7
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

解法：使用DFS

  • DFS方法： 设目前指针指向一个岛屿中的某一点 (i, j)，寻找包括此点的岛屿边界。
    • 从 (i, j) 向此点的上下左右 (i+1,j),(i-1,j),(i,j+1),(i,j-1) 做深度搜索。
    • 终止条件：
      • (i, j) 越过矩阵边界;
      • grid[i][j] == 0，代表此分支已越过岛屿边界。
    • 搜索岛屿的同时，执行 grid[i][j] = ‘0’，即将岛屿所有节点删除，以免之后重复搜索相同岛屿。
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
class Solution(object):
    def numIslands(self, grid):
        """
        :type grid: List[List[str]]
        :rtype: int
        """
        def dfs(grid, 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(grid,i+1,j)
            dfs(grid,i,j+1)
            dfs(grid,i-1,j)
            dfs(grid,i,j-1)
        count = 0
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if grid[i][j] == '1':
                    dfs(grid,i,j)
                    count += 1
        return count