[Union Find] Add a solution to Number of Islands II

Author: soapyigu

README.md

@@ -28,7 +28,7 @@
 * [Microsoft](#microsoft)
 
 ## Progress
-[Problem Status](#problem-status) shows the latest progress to all 800+ questions. Currently we have 237 completed solutions. Note: questions with ♥ mark means that you have to **Subscript to premium membership** of LeetCode to unlock them. Thank you for great contributions from [CharleneJiang](https://github.com/CharleneJiang), [ReadmeCritic](https://github.com/ReadmeCritic), [demonkoo](https://github.com/demonkoo), [DaiYue](https://github.com/DaiYue), [Quaggie](https://github.com/Quaggie) and [jindulys](https://github.com/jindulys).
+[Problem Status](#problem-status) shows the latest progress to all 800+ questions. Currently we have 238 completed solutions. Note: questions with ♥ mark means that you have to **Subscript to premium membership** of LeetCode to unlock them. Thank you for great contributions from [CharleneJiang](https://github.com/CharleneJiang), [ReadmeCritic](https://github.com/ReadmeCritic), [demonkoo](https://github.com/demonkoo), [DaiYue](https://github.com/DaiYue), [Quaggie](https://github.com/Quaggie) and [jindulys](https://github.com/jindulys).
 
 
 ## Array
@@ -305,6 +305,7 @@
 | ----- | -------- | ---------- | ---- | ----- |
 [Number of Connected Components in an Undirected Graph](https://leetcode.com/problems/number-of-connected-components-in-an-undirected-graph/)| [Swift](./UnionFind/NumberConnectedComponentsUndirectedGraph.swift)| Medium| O(nlogn)| O(n)|
 [Graph Valid Tree](https://leetcode.com/problems/graph-valid-tree/)| [Swift](./UnionFind/GraphValidTree.swift)| Medium| O(nlogn)| O(n)|
+[Number of Islands II](https://leetcode.com/problems/number-of-islands-ii/)| [Swift](./UnionFind/NumberIslandsII.swift)| Hard| O(klogmn)| O(mn)|
 
 ## Google
 | Title | Solution | Difficulty | Frequency |
@@ -507,7 +508,7 @@
 |		|	308	|	[Range Sum Query 2D - Mutable](https://leetcode.com/problems/range-sum-query-2d-mutable/) ♥	|	Hard
 |		|	307	|	[Range Sum Query - Mutable](https://leetcode.com/problems/range-sum-query-mutable/)	|	Medium
 |		|	306	|	[Additive Number](https://leetcode.com/problems/additive-number/)	|	Medium
-|		|	305	|	[Number of Islands II](https://leetcode.com/problems/number-of-islands-ii/)  ♥	|	Hard
+|	[Swift](./UnionFind/NumberIslandsII.swift)	|	305	|	[Number of Islands II](https://leetcode.com/problems/number-of-islands-ii/)  ♥	|	Hard
 |		|	304	|	[Range Sum Query 2D - Immutable](https://leetcode.com/problems/range-sum-query-2d-immutable/)	|	Medium
 |		|	303	|	[Range Sum Query - Immutable](https://leetcode.com/problems/range-sum-query-immutable/)	| Easy
 |		|	302	|	[Smallest Rectangle Enclosing Black Pixels](https://leetcode.com/problems/smallest-rectangle-enclosing-black-pixels/)  ♥	|	Hard

UnionFind/NumberIslandsII.swift

@@ -0,0 +1,48 @@
+/**
+ * Question Link: https://leetcode.com/problems/number-of-islands-ii/
+ * Primary idea: Classic Union Find, check four directions and update count every time
+ * 
+ * Time Complexity: O(klogmn), Space Complexity: O(mn)
+ *
+ */
+
+class NumberIslandsII {
+  func numOfIslandsII(_ m: Int, _ n: Int, _ positions: [(Int, Int)]) -> [Int] {
+    var res = [Int](), count = 0, roots = Array(repeating: -1, count: m * n)
+    
+    for position in positions {
+      var pos = position.0 * n + position.1
+      roots[pos] = pos
+      count += 1
+      
+      for moveDir in [(0, 1), (0, -1), (1, 0), (-1, 0)] {
+        let i = position.0 + moveDir.0, j = position.1 + moveDir.1
+        let movePos = i * n + j
+        
+        guard i >= 0 && i < m && j >= 0 && j < n && roots[movePos] != -1 else {
+          continue
+        }
+        
+        let movePosRoot = findRoot(movePos, roots)
+        
+        if movePosRoot != pos {
+          count -= 1
+          roots[pos] = movePosRoot
+          pos = movePosRoot
+        }
+      }
+      
+      res.append(count)
+    }
+    
+    return res
+  }
+
+  fileprivate func findRoot(_ node: Int, _ roots: [Int]) -> Int {
+    var node = node
+    while node != roots[node] {
+      node = roots[node]
+    }
+    return node
+  }
+}