理论基础：

 

第一题、所有可能的路径 力扣题目链接

class Solution {
private:
    vector<vector<int>> result;
    vector<int> path;

    void dfs(vector<vector<int>>& graph, int x){
        if(x == graph.size() - 1){
            result.push_back(path);
            return;
        }
        for(int i = 0; i < graph[x].size(); i++){
            path.push_back(graph[x][i]);
            dfs(graph, graph[x][i]);
            path.pop_back();
        }
    }
public:
    vector<vector<int>> allPathsSourceTarget(vector<vector<int>>& graph) {
        path.push_back(0);
        dfs(graph, 0);
        return result;
    }
};

第二题、岛屿数量 力扣题目链接

深搜版：

class Solution {
private:
    int dir[4][2] = {0, 1, 1, 0, -1, 0, 0, -1};  // 加减1控制4个方向
    void dfs(vector<vector<char>>& grid, vector<vector<bool>>& visited, int x, int y){
        for(int i = 0; i < 4; i++){
            int nextx = x + dir[i][0];
            int nexty = y + dir[i][1];
            if(nextx < 0 || nextx >= grid.size() || nexty < 0 || nexty >= grid[0].size()) continue;
            if(!visited[nextx][nexty] && grid[nextx][nexty] == '1'){
                visited[nextx][nexty] = true;
                dfs(grid, visited, nextx, nexty);
            }
        }
    }
public:
    int numIslands(vector<vector<char>>& grid) {
        int n = grid.size(), m = grid[0].size();
        vector<vector<bool>> visited = vector<vector<bool>>(n, vector<bool>(m, false));
        int result = 0;
        for(int i = 0; i < n; i++){
            for(int j = 0; j < m; j++){
                if(!visited[i][j] && grid[i][j] == '1'){
                    visited[i][j] = true;
                    result++;
                    dfs(grid, visited, i, j);
                }
            }
        }
        return result;
    }
};

广搜版：

class Solution {
private:
    int dir[4][2] = {0, 1, 1, 0, -1, 0, 0, -1};  // 加减1控制4个方向
    void bfs(vector<vector<char>>& grid, vector<vector<bool>>& visited, int x, int y){
        queue<pair<int, int>> que;
        que.push({x, y});
        visited[x][y] = true;  // 立刻标记
        while(!que.empty()){
            pair<int, int> cur = que.front(); que.pop();
            int curx = cur.first;
            int cury = cur.second;
            for(int i = 0;i < 4; i++){
                int nextx = curx + dir[i][0];
                int nexty = cury + dir[i][1];
                if(nextx < 0 || nextx >= grid.size() || nexty < 0 || nexty >= grid[0].size()) continue;
                if(!visited[nextx][nexty] && grid[nextx][nexty] == '1'){
                    que.push({nextx, nexty});
                    visited[nextx][nexty] = true;  // 加入队列 立刻标记
                }
            }
        }
    }
public:
    int numIslands(vector<vector<char>>& grid) {
        int n = grid.size(), m = grid[0].size();
        vector<vector<bool>> visited = vector<vector<bool>>(n, vector<bool>(m, false));

        int result = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (!visited[i][j] && grid[i][j] == '1') {
                    result++; // 遇到没访问过的陆地，+1
                    bfs(grid, visited, i, j); // 将与其链接的陆地都标记上 true
                }
            }
        }
        return result;
    }
};