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，+1bfs(grid, visited, i, j); // 将与其链接的陆地都标记上 true}}}return result;}
};