1. 题目

节点间通路。给定有向图，设计一个算法，找出两个节点之间是否存在一条路径。

示例1:输入：n = 3, graph = [[0, 1], [0, 2], [1, 2], [1, 2]],start = 0, target = 2输出：true
示例2:输入：n = 5, graph = [[0, 1], [0, 2], [0, 4], [0, 4], [0, 1], [1, 3], [1, 4], [1, 3], [2, 3], [3, 4]], start = 0, target = 4输出 true提示：
节点数量n在[0, 1e5]范围内。
节点编号大于等于 0 小于 n。
图中可能存在自环和平行边。

2. 解题

• 邻接矩阵 内存超限

class Solution {
public:bool findWhetherExistsPath(int n, vector<vector<int>>& graph, int start, int target) {bool visited[n] = {false};visited[start] = true;queue<int> q;int tp, i;q.push(start);vector<vector<int>> map(n,vector<int>(n,0));for(i = 0; i < graph.size(); ++i)map[graph[i][0]][graph[i][1]] = 1;while(!q.empty()){tp = q.front();q.pop();if(tp == target)return true;for(i = 0; i < n; ++i){if(!visited[i] && map[tp][i]==1){visited[i] = true;q.push(i);}}}return false;}
};

• 邻接表 BFS

class Solution {
public:bool findWhetherExistsPath(int n, vector<vector<int>>& graph, int start, int target) {bool visited[n] = {false};visited[start] = true;vector<vector<int>> map(n);int i, tp;for(i = 0; i < graph.size(); ++i){map[graph[i][0]].push_back(graph[i][1]);}queue<int> q;q.push(start);while(!q.empty()){tp = q.front();if(tp == target)return true;q.pop();for(i = 0; i < map[tp].size(); ++i){if(!visited[map[tp][i]]){q.push(map[tp][i]);visited[map[tp][i]] = true;}}}return false;}
};

• 邻接表 DFS

class Solution {
public:bool findWhetherExistsPath(int n, vector<vector<int>>& graph, int start, int target) {bool visited[n] = {false};visited[start] = true;vector<vector<int>> map(n);for(int i = 0; i < graph.size(); ++i){map[graph[i][0]].push_back(graph[i][1]);}bool found = false;dfs(map,start,target,found,visited);return found;}void dfs(vector<vector<int>>& map , int start, int& target, bool& found, bool* visited){if(start == target)found = true;if(found)return;for(int i = 0; i < map[start].size(); ++i){if(!visited[map[start][i]]){visited[map[start][i]] = true;dfs(map,map[start][i],target,found,visited);visited[map[start][i]] = false;}}}
};