Hexo

You are given an array points representing integer coordinates of some points on a 2D-plane, where points[i] = [xi, yi].

The cost of connecting two points [xi, yi] and [xj, yj] is the manhattan distance between them: |xi - xj| + |yi - yj|, where |val| denotes the absolute value of val.

Return the minimum cost to make all points connected. All points are connected if there is exactly one simple path between any two points.

并查集，最小生成树(Minimum spanning tree)，Kruskal算法。
Edge辅助类，保存并计算两点的id和其曼哈顿距离。
将edges根据其距离排序。

遍历所有的edge，如果edge的两点没有合并，则合并两点，并取这条边作为结果。
加和所有的结果，答案就是最小值。

class Solution {
    public int minCostConnectPoints(int[][] points) {
        List<Edge> edges = new ArrayList<Edge>();
        int min = 0;
        for(int i = 0; i < points.length; i++){
            for(int j = i+1; j < points.length; j++){
                 edges.add(new Edge(points, i, j));
            }
        }
        
        Collections.sort(edges, (a, b) -> a.length - b.length);
        UnionFind uf = new UnionFind(points.length);
        
        for(Edge e : edges){
            if(uf.union(e.id1, e.id2)){
                min += e.length;
            }
        }
        return min;
    }

    class UnionFind{
        private int[] parents;
        private int[] size;
    
        public UnionFind(int n){
            size = new int[n];
            parents = new int[n];
            for(int i = 0; i < n; i++){
                parents[i] = i;
                size[i] = 1;
            }
        }
    
        public int find(int id){
            if(parents[id] == id) return id;
            return find(parents[id]);
        }
    
        public boolean union(int id1, int id2){
            int p1 = find(id1);
            int p2 = find(id2);
            
            if(p1 == p2) return false;
        
            if(size[p1] < size[p2]){
                parents[p1] = p2;
                size[p2] += size[p1];
            }
            else{
                parents[p2] = p1;
                size[p1] += size[p2];
            }
            return true;
        }
    }

    class Edge{
        int id1;
        int id2;
        int length;
        public Edge(int[][] points, int _id1, int _id2){
            id1 = _id1;
            id2 = _id2;
            length = Math.abs(points[id1][0] - points[id2][0]) + Math.abs(points[id1][1] - points[id2][1]);
        }
    }
}