number of islands using union find

Author: fishercoder1534

MEDIUM/src/medium/NumberOfIslandsUnionFind.java

@@ -0,0 +1,66 @@
+package medium;
+
+/**
+ * Created by fishercoder1534 on 9/30/16.
+ */
+
+//Inspired by this post: https://discuss.leetcode.com/topic/39980/1d-union-find-java-solution-easily-generalized-to-other-problems
+public class NumberOfIslandsUnionFind {
+
+    class UnionFind{
+        int count;
+        int m, n;
+        int[] ids;
+
+        public UnionFind(char[][] grid){
+            m = grid.length;
+            n = grid[0].length;
+            ids = new int[m*n];
+            for(int i = 0; i < m; i++){
+                for(int j = 0; j < n; j++){
+                    if(grid[i][j] == '1') {
+                        count++;
+                        ids[i*n+j] = i*n+j;
+                    }
+                }
+            }
+        }
+
+        public void union(int i, int j){
+            int x = find(ids, i);
+            int y = find(ids, j);
+            if(x != y) {//note: this is when x != y, only in this case, we should union these two nodes, which makes sense naturally.
+                count--;
+                ids[x] = y;
+            }
+        }
+
+        public int find(int[] ids, int i){
+            if(ids[i] == i) return i;
+            return find(ids, ids[i]);
+        }
+    }
+
+    public int numIslands(char[][] grid) {
+        if(grid == null || grid.length == 0 || grid[0].length == 0) return 0;
+        int[] dirs = new int[]{0,1,0,-1,0};
+        UnionFind uf = new UnionFind(grid);
+        int m = grid.length, n = grid[0].length;
+        for(int i = 0; i < m; i++){
+            for(int j = 0; j < n; j++){
+                if(grid[i][j] == '1'){
+                    for(int k = 0; k < 4; k++){
+                        int x = i+dirs[k];
+                        int y = j+dirs[k+1];
+                        if(x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == '1'){
+                            int id1 = i*n+j;
+                            int id2 = x*n+y;
+                            uf.union(id1, id2);
+                        }
+                    }
+                }
+            }
+        }
+        return uf.count;
+    }
+}