Graph Valid Tree

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主要是看了这篇文章

http://www.geeksforgeeks.org/union-find/

总结下目前理解的Union-Find

用在disjoint-set data structure上，disjoint-set keeps track of a set of elements partitioned into a number of disjoint (non-overlapping) subsets.

Union-Find is an algorithm that performs two useful operations on such a data structure:

1. Find

determine which subset a particular element is in, this can be used to determin whether two elements are in the same subset

2. Union

join two subsets into a single one

Union-Find can be used to check whether an undirected graph contains cycle or not, and this algorithm assumes that graph doesn‘t contain any self-loops

say we have a graph as below:

[0, 1]

[1, 2]

[2, 0]

When using union-find to process the graph, we use a hashmap to store the parent-child relationship. 

1) first we have node 0 and node 1

the map doesn‘t contain 0 nor 1, so it returns 0 and 1 themselves (because they are the parent of themselves by now)

And we connect them together by making 0 the parent of 1 or 1 the parent of 0

so put (0->1) in the map

2) and then we have node 1 and node 2

the map doesn‘t contain 1 nor 2 so it returns 1 for node 1 and returns 2 for node 2

And we connect them together by putting (1->2) in the map

now we have (0->1), (1->2) in the map

3) For node 2 and node 0

the parent of node 2 is 2 itself, but the parent of node 0 is also 2, they equal to each other meaning that they belong to the same subset

Therefore, we found a cycle

In the code, I use an array instead of a hashmap, and fill the array with -1 at first meaning that this is the end, this number is the parent of itself

    public boolean validTree(int n, int[][] edges) {
        if(edges == null) return false;
        int[] parent = new int[n];
        Arrays.fill(parent, -1);
        
        for(int i = 0; i < edges.length; i++){
            int x = find(parent, edges[i][0]);
            int y = find(parent, edges[i][1]);
            if(x==y) return false;
            
            parent[x] = y;
        }
        
        return edges.length == n-1;
    }
    
    private int find(int[] parent, int i){
        if(parent[i] == -1) return i;
        return find(parent, parent[i]);
    }

也可以用在另外两道lintcode题目上 

Find the Weak Connected Component in the Directed Graph

Find the Connected Component in the Undirected Graph

代码类似，总的来说union-find就是在disjoint-set 的题目上面很有帮助

Graph Valid Tree

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原文地址：http://www.cnblogs.com/momoco/p/4845413.html