200. 岛屿数量

题目描述

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'

解答

class Solution {
public:
    int numIslands(vector<vector<char>>& grid) {
        // 对每一个点进行dfs
        int res = 0;
        int row = grid.size(), col = grid[0].size();
        for(int i = 0; i < row; ++i)
        {
            for(int j = 0; j < col; ++j)
            {
                if(grid[i][j] == '1')
                {
                    ++res;
                    dfs(grid, i, j); // 从 grid(i, j) 开始dfs, 所有经过的位置打上标记,即把整座岛屿标记为访问过为2
                }
            }
        }
        return res;
    }
    void dfs(vector<vector<char>> &grid, int x, int y)
    {
        if(x < 0 || x >= grid.size() || y < 0 || y >= grid[0].size() || grid[x][y] != '1')
        {
            return ;
        }
        grid[x][y] = '2';
        
        dfs(grid, x - 1, y);
        dfs(grid, x + 1, y);
        dfs(grid, x, y - 1);
        dfs(grid, x, y + 1);
    }
};