Leetcode
Graph
Depth-first Search
Breath-first Search
Topological Sort
There are a total of n courses you have to take, labeled from 0 to n-1.
Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1]
Given the total number of courses and a list of prerequisite pairs, is it possible for you to finish all courses?
Example 1:
Input: 2, [[1,0]] Output: true Explanation: There are a total of 2 courses to take. To take course 1 you should have finished course 0. So it is possible.
Example 2:
Input: 2, [[1,0],[0,1]] Output: false Explanation: There are a total of 2 courses to take. To take course 1 you should have finished course 0, and to take course 0 you should also have finished course 1. So it is impossible.
Note:
- The input prerequisites is a graph represented by a list of edges, not adjacency matrices.
- You may assume that there are no duplicate edges in the input prerequisites.
分析¶
DFS的常见应用。检查有向图是不是DAG，即有没有环。详见Algorithms 4th，思路很简单，如果在遍历了v->w时，发现w在栈上，那么说明肯定有环，因为既然w在栈上，说明肯定有一条路径使得w->v。
private boolean hasCycle; public boolean canFinish(int numCourses, int[][] prerequisites) { hasCycle = false; // construct a graph List<List<Integer>> graph = new ArrayList<>(); for (int i = 0; i < numCourses; i++) graph.add(new ArrayList<Integer>()); for (int[] prerequisite : prerequisites) graph.get(prerequisite[1]).add(prerequisite[0]); // dfs boolean[] mark = new boolean[numCourses]; boolean[] onStack = new boolean[numCourses]; for (int v = 0; v < numCourses; v++) if (!hasCycle & !mark[v]) dfs(graph, mark, onStack, v); return !hasCycle; } private void dfs(List<List<Integer>> graph, boolean[] mark, boolean[] onStack, int v) { mark[v] = true; onStack[v] = true; for (int w : graph.get(v)) { if (!hasCycle & !mark[w]) dfs(graph, mark, onStack, w); else if (onStack[w]) hasCycle = true; } onStack[v] = false; }
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