所有可能的路径：

力扣（LeetCode）官网 - 全球极客挚爱的技术成长平台 

 

class Solution {
public:
    vector<int> path;
    vector<vector<int>> result;
    void dfs(vector<vector<int>> graph, int x){//1 确认递归函数和参数
        if (x == graph.size()-1) {//2 确认终止条件
            result.push_back(path);
            return;
        }

        for (int i = 0; i < graph[x].size(); i++) {//3 处理目前搜索节点出发的路径
            path.push_back(graph[x][i]);//把当前节点压入
            dfs(graph, graph[x][i]);//递归
            path.pop_back();//回溯
        }
    }

    vector<vector<int>> allPathsSourceTarget(vector<vector<int>>& graph) {
        path.push_back(0);//初始节点0要加入
        dfs(graph, 0);//从初始节点开始递归
        return result;
    }
};

bfs：

用队列实现：

岛屿数量

力扣（LeetCode）官网 - 全球极客挚爱的技术成长平台

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<vector<>> 
     * @return int整型
     */
    int dir[4][2] = {1,0, 0,1, -1,0, 0,-1};
    void dfs(vector<vector<char>>& grid, vector<vector<bool>>& visited, int x, int y) {
        for (int i = 0; i < 4; i++) {
            int newX = x + dir[i][0];
            int newY = y + dir[i][1];
            if (newX < 0 || newX >= grid.size() || newY < 0 || newY >= grid[0].size()) continue;
            if (!visited[newX][newY] && grid[newX][newY] == '1') {
                visited[newX][newY] = true;
                dfs(grid, visited, newX, newY);
            }
        }
    }

    int solve(vector<vector<char> >& grid) {
        // write code here
        vector<vector<bool>> visited(grid.size(), vector<bool> (grid[0].size(), false));
        int count = 0;
        for (int i = 0; i < grid.size(); i++) {
            for (int j = 0; j < grid[0].size(); j++) {
                if (!visited[i][j] && grid[i][j] == '1') {
                    count++;
                    visited[i][j] = true;
                    dfs(grid, visited, i, j);
                }
            }
        }
        return count;
    }
};

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 dir[4][2] = {1,0, 0,1, -1,0, 0,-1};
    void bfs(vector<vector<char>>& grid, vector<vector<bool>>& visited, int x,int y) {
        queue<pair<int, int>> que;
        que.push(pair(x, y));
        visited[x][y] = true;
        while (!que.empty()) {
            int curX = que.front().first;
            int curY = que.front().second;
            que.pop();
            for(int i = 0; i < 4; i++) {
                int newX = curX + dir[i][0];
                int newY = curY + dir[i][1];
                if (newX < 0 || newX >= grid.size() || newY < 0 || newY >= grid[0].size()) continue;
                if (!visited[newX][newY] && grid[newX][newY] == '1') {
                    que.push(pair(newX, newY));
                    visited[newX][newY] = true;
                }
            }
        }
    }

    int solve(vector<vector<char> >& grid) {
        // write code here
        vector<vector<bool>> visited(grid.size(), vector<bool> (grid[0].size(), false));
        int count = 0;
        for (int i = 0; i < grid.size(); i++) {
            for (int j = 0; j < grid[0].size(); j++) {
                if (!visited[i][j] && grid[i][j] == '1') {
                    count++;
                    bfs(grid, visited, i, j);
                }
            }
        }
        return count;

    }
};