https://www.nowcoder.com/practice/0c9664d1554e466aa107d899418e814e?tpId=196&tqId=37167&ru=/exam/oj

dfs

#include <vector>
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 判断岛屿数量
     * @param grid char字符型vector<vectovecr<>> 
     * @return int整型
     */


    vector<vector<int>> vv; // 记录是否被遍历过
    int dx[4] = {0, 0, 1, -1};
    int dy[4] = {1, -1, 0, 0};
    int m, n;
    int solve(vector<vector<char> >& grid) {
        // write code here
        m = grid.size();
        n = grid[0].size();
        vv.assign(m, vector<int>(n, 0));
        int ret = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == '1' && !vv[i][j]) {
                    //cout << i << "  " << j << endl;
                    dfs(grid, i, j);
                    ret++;
                }
            }
        }
        return ret;
    }

    void dfs(vector<vector<char> >& grid, int i, int j) {
        for (int k = 0; k < 4; ++k) {
            int a = i + dx[k], b = j + dy[k];
            if (a >= 0 && a < m && b >= 0 && b < n && grid[a][b] == '1' && !vv[a][b]) {
                //cout << a << "  " << b << endl;
                vv[a][b] = 1;
                dfs(grid, a, b);
            }
        } 
    }
};

bfs:

class Solution {
  public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 判断岛屿数量
     * @param grid char字符型vector<vector<>>
     * @return int整型
     */

    int n, m;
    int dx[4] = {0, 0, 1, -1};
    int dy[4] = {1, -1, 0, 0};
    vector<vector<bool>> vv;

    int solve(vector<vector<char> >& grid) {
        // write code here
        m = grid.size(), n = grid[0].size();
        vv.assign(m, vector<bool>(n, false));

        int ret = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == '1' && !vv[i][j]) {
                    //cout << i << "  " << j << endl;
                    bfs(grid, i, j);
                    ret++;
                }
            }
        }
        return ret;
    }

    void bfs(vector<vector<char> >& grid, int i, int j) {
        queue<pair<int, int>> q;
        q.push({i, j});
        vv[i][j] = true;

        while (!q.empty()) {
            auto [x, y] = q.front();
            q.pop();

            for (int k = 0; k < 4; ++k) {
                int a = x + dx[k], b = y + dy[k];
                if (a >= 0 && a < m && b >= 0 && b < n && grid[a][b] == '1' && !vv[a][b]) {
                    q.push({a, b});
                    vv[a][b] = true;
                }
            }
        }
    }
};