Java for LeetCode 207 Course Schedule 【Unsolved】

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?

For example:

2, [[1,0]]

There are a total of 2 courses to take. To take course 1 you should have finished course 0. So it is possible.

2, [[1,0],[0,1]]

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.

解题思路一：

把每个节点放到图里，然后对每个节点进行BFS，如果出现自环，则为false，否则返回true，JAVA实现如下：

static public boolean canFinish(int numCourses, int[][] prerequisites) {
    if (prerequisites.length == 0 || prerequisites[0].length == 0)
        return true;
    HashMap<Integer, UndirectedGraphNode> hm = new HashMap<Integer, UndirectedGraphNode>();
    for (int[] nums : prerequisites)
        for (int i = 0; i < nums.length - 1; i++) {
            if (!hm.containsKey(nums[i]))
                hm.put(nums[i], new UndirectedGraphNode(nums[i]));
            if (!hm.containsKey(nums[i + 1]))
                hm.put(nums[i + 1], new UndirectedGraphNode(nums[i + 1]));
            hm.get(nums[i]).neighbors.add(hm.get(nums[i + 1]));
        }
    Iterator<Integer> iterator = hm.keySet().iterator();
    while (iterator.hasNext())
        if (haveLoop(hm.get(iterator.next())))
            return false;
    return true;
}

static boolean haveLoop(UndirectedGraphNode root) {
	HashMap<UndirectedGraphNode, Boolean> hm = new HashMap<UndirectedGraphNode, Boolean>();
	Queue<UndirectedGraphNode> queue = new LinkedList<UndirectedGraphNode>();
	hm.put(root, true);
	queue.add(root);
	while (!queue.isEmpty()) {
		UndirectedGraphNode temp = queue.poll();
		for (UndirectedGraphNode temp2 : temp.neighbors) {
			if (temp2 == root)
				return true;
			if (!hm.containsKey(temp2)) {
				hm.put(temp2, true);
				queue.add(temp2);
			}
		}
	}
	return false;
}

结果TLE，问题在于判断有向图是否有环的时候，对每一个节点判断其实浪费了很多时间%>_<%