Leetcod graph(new topic)

797. All Paths From Source to Target

class Solution {
private:
    vectorint>> res;
    void traversal(vectorint>>& graph, vector<int>& path, int s){
        path.push_back(s);
        int n=graph.size();
        if(s == n-1){
            res.push_back(path);
        }
        for(auto v:graph[s]){
            traversal(graph, path, v);
        }
        path.pop_back();
    }
public:
    vectorint>> allPathsSourceTarget(vectorint>>& graph) {
        vector<int> path;
        traversal(graph, path, 0);
        return res;

类似于多叉树

这里的s是记录正在走的节点

如果不是acyclic(无环), 要用另外一个vector 记录走过的节点，防止进入死循环

277. Find the Celebrity

class Solution {
public:
    int findCelebrity(int n) {
        int cand = 0;
        for(int other = 1; other < n; other++){
            if(knows(cand, other) || !knows(other, cand)){
                cand = other;
            }
        }
 
        for(int other = 0; other
            if(cand == other) continue;
            if(knows(cand, other) || !knows(other, cand)) return -1;
        }
        return cand;
    }
};

1.首先要排除，因为条件，所以至多有一个名人，那么两两对比，选出唯一的可能是名人的那个人

2.对比的结果有四种

3.最后剩下的那一个也不一定是名人，要在判断

207. Course Schedule

1.DFS

class Solution {
private:
    bool hasCycle = false;
    vector<bool> onPath;
    vector<bool> visited;
    vectorint>> buildGraph(int numCourses, vectorint>>& prerequisites){
        vectorint>> graph(numCourses);
        for(auto prerequisite:prerequisites){
            int from = prerequisite[1];
            int to = prerequisite[0];
            graph[from].push_back(to);
        }
        return graph;
    }
    void traversal(vectorint>>& graph, int s){
        if(onPath[s]){
            hasCycle = true;
        }
        if(visited[s] || hasCycle){
            return;
        }
        visited[s] = true;
        onPath[s] = true;
        for(auto t:graph[s]){
            traversal(graph, t);
        }
        onPath[s] = false;
    }
public:
    bool canFinish(int numCourses, vectorint>>& prerequisites) {
        vectorint>> graph = buildGraph(numCourses, prerequisites);
        visited = vector<bool>(numCourses);
        onPath = vector<bool>(numCourses);
        for(int i=0; i
            traversal(graph, i);
        }
        return !hasCycle;
    }
};

1.首先要建一个graph表

2.遍历判断是否有环

210. Course Schedule II

class Solution {
private:
    vector<int> res;
    bool hasCycle;
    vector<bool> visited, onPath;
 
    void traversal(vectorint>>& graph, int s){
        if(onPath[s]){
            hasCycle = true;
        }
        if(visited[s] || hasCycle) return;
        onPath[s] = true;
        visited[s] = true;
 
        for(int t:graph[s]){
            traversal(graph, t);
        }
        res.push_back(s);
        onPath[s] = false;
    }
 
    vectorint>> buildGraph(int numCourses, vectorint>>& prerequisites){
        vectorint>> graph(numCourses);
        for(auto prerequisite:prerequisites){
            int from = prerequisite[1];
            int to = prerequisite[0];
            graph[from].push_back(to);
        }
        return graph;
    }
public:
    vector<int> findOrder(int numCourses, vectorint>>& prerequisites) {
        vectorint>> graph = buildGraph(numCourses, prerequisites);
        visited = vector<bool>(numCourses, false);
        onPath = vector<bool>(numCourses, false);
        for(int i=0; i
            traversal(graph, i);
        }
        if(hasCycle) return {};
        reverse(res.begin(), res.end());
        return res;
    }
};

需要一个单独的vector记录

这里用的是后序，所以之后要reverse一下