M200.岛屿数量

dfs, bfs, https://leetcode.cn/problems/number-of-islands/

思路：通过dfs，bfs遍历算法，在遇到陆地('1')时进行连通分量标记并计数，利用搜索过程将相邻陆地沉没为水域从而避免重复统计。

class Solution {
public:
    int numIslands(vector<vector<char>>& grid) {
        if (grid.empty()) return 0; 
        int m = grid.size();     
        int n = grid[0].size();     
        int count = 0;             
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == '1') { 
                    count++;             
                    dfs(grid, i, j, m, n); 
                }
            }
        }
        return count;
    }

private:
    void dfs(vector<vector<char>>& grid, int i, int j, int m, int n) {
        if (i < 0 || i >= m || j < 0 || j >= n || grid[i][j] != '1') {
            return;
        }
        grid[i][j] = '0'; 
        dfs(grid, i - 1, j, m, n);
        dfs(grid, i + 1, j, m, n); 
        dfs(grid, i, j - 1, m, n); 
        dfs(grid, i, j + 1, m, n); 
    }
};