图论第一天|797.所有可能的路径 200. 岛屿数量

本文主要是介绍图论第一天|797.所有可能的路径 200. 岛屿数量，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

Leetcode797.所有可能的路径

文章链接：代码随想录
题目链接：797.所有可能的路径

思路：深搜入门，注意邻接表和邻接矩阵的形式

class Solution {
public: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();}}vector<vector<int>> allPathsSourceTarget(vector<vector<int>>& graph) {path.push_back(0);dfs(graph, 0);return result;}
};

Leetcode200. 岛屿数量

文章链接：代码随想录
题目链接：200. 岛屿数量

思路：深搜法

class Solution {
public:int dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};void dfs(vector<vector<char>>& grid, vector<vector<bool>>& visited, int x, int y){for (int i = 0; i < 4; i++){int nex = x + dir[i][0];int ney = y + dir[i][1];if (nex < 0 || nex >= grid.size() || ney < 0 || ney >= grid[0].size()) continue ;if (!visited[nex][ney] && grid[nex][ney] == '1'){visited[nex][ney] = true;dfs(grid, visited, nex, ney);}}}int numIslands(vector<vector<char>>& grid) {int result = 0;vector<vector<bool>> visited(grid.size(), vector<bool>(grid[0].size(), 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'){result++;visited[i][j] = true;dfs(grid, visited, i, j);}}}return result;}
};

广搜法，用队列存储，遍序搜寻，替代深搜回溯

class Solution {
public:int dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};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;while(!que.empty()){pair<int, int> cur = que.front();que.pop();for (int i = 0; i < 4; i++){int nex = cur.first + dir[i][0];int ney = cur.second + dir[i][1];if (nex < 0 || nex >= grid.size() || ney < 0 || ney >= grid[0].size()) continue ;if (!visited[nex][ney] && grid[nex][ney] == '1'){que.push({nex, ney});visited[nex][ney] = true;}}}}int numIslands(vector<vector<char>>& grid) {int result = 0;vector<vector<bool>> visited(grid.size(), vector<bool>(grid[0].size(), 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'){result++;bfs(grid, visited, i, j);}}}return result;}
};