【leetcode】684. Redundant Connection

可以用递归或者union set

union set

class Solution {
 	public int[] findRedundantConnection(int[][] edges) {
    	int[] us=new int[2000];
    	for(int i=0;i<2000;i++)
    		us[i]=i;
    	
    	for(int[] e:edges) {
    		if(findSet(e[0], us)==findSet(e[1], us))
    			return e;
    		union(e[0],e[1], us);
    	}
    	
    	return null;
    }
 	
 	private int findSet(int num, int[] us) {
 		if(num==us[num])		return num;
 		while(us[num]!=num)
 			num=us[num];
 		return num;
 	}
 	
 	private void union(int start, int end, int[] us) {
 		int startSet = findSet(start, us);
 		int endSet = findSet(end, us);
 		if(startSet!=endSet) {
 			us[startSet]=endSet;
 		}
 	}
 	
}

运行结果

递归

class Solution {
	Map<Integer, List<Integer>> graph=new HashMap<>();
	Set<Integer> allNodes = new HashSet<>();
	
 	public int[] findRedundantConnection(int[][] edges) {
    	for(int[] e: edges) {
    		if(findPath(e[0], e[1]))
    			return e;
    		graph.computeIfAbsent(e[0], k->new ArrayList<>()).add(e[1]);
    		graph.computeIfAbsent(e[1], k->new ArrayList<>()).add(e[0]);
    	}
        return null;
    }
 	
 	private Boolean findPath(int start, int end) {
 		if(!graph.containsKey(start)||!graph.containsKey(end))	return false;
 		for(int i=0; i<graph.get(start).size();i++) {
 			int curNode = graph.get(start).get(i);
 			if(allNodes.contains(curNode))	continue;
 			allNodes.add(curNode);
 			if(curNode == end)	return true;
 			if(findPath(curNode, end))	return true;
 			allNodes.remove(curNode);
 		}
 		return false;
 	}
}

运行结果