There is a list composed by sets. If two sets have the same elements, merge them. In the end, there are several sets left.

Example

Example 1:

Input :list = [[1,2,3],[3,9,7],[4,5,10]]
Output:2 .
Explanation:There are 2 sets of [1,2,3,9,7] and [4,5,10] left.

Example 2:

Input:list = [[1],[1,2,3],[4],[8,7,4,5]]
Output :2
Explanation:There are 2 sets of [1,2,3] and [4,5,7,8] left.

Notice

• The number of sets n <=1000.
• The number of elements for each set m <= 100.
• The element must be a non negative integer and not greater than 100000.

思路：Union Find的模板，先find最大值，然后申请union Find数组，然后用hashset统计一下点的个数，然后union一下，这里取巧，union的时候return一下true和false，这样count可以--；

public class Solution {
    /**
     * @param sets: Initial set list
     * @return: The final number of sets
     */
    private class UnionFind{
        private int[] father;
        private int count;
        
        public UnionFind(int n) {
            father = new int[n+1];
            for(int i = 0; i <=n; i++) {
                father[i] = i;
            }
            this.count = n;
        }
        
        public int find(int x) {
            int j = x;
            while(father[j] != j) {
                j = father[j];
            }
            
            // path compression
            while(x != j) {
                int fx = father[x];
                father[x] = j;
                x = fx;
            }
            
            return j;
        }
        
        public boolean union(int a, int b) {
            int father_a = find(a);
            int father_b = find(b);
            if(father_a != father_b) {
                father[father_a] = father_b;
                return true;
            }
            return false;
        }
    }
     
    public int setUnion(int[][] sets) {
        if(sets == null || sets.length == 0) {
            return 0;
        }
        HashSet<Integer> hashset = new HashSet<Integer>();
        int max = Integer.MIN_VALUE;
        for(int i = 0; i < sets.length; i++) {
            for(int j = 0; j < sets[i].length; j++) {
                max = Math.max(max, sets[i][j]);
                hashset.add(sets[i][j]);
            }
        }
        
        UnionFind uf = new UnionFind(max);
        int count = hashset.size();
        for(int i = 0; i < sets.length; i++) {
            for(int j = 0; j < sets[i].length-1; j++) {
                if(uf.union(sets[i][j], sets[i][j+1])) {
                    count--;
                }
            }
        }
        return count;
    }
}