[Search] Add solutions to Search 2D Matrix I and II

Author: Yi Gu

Diff for: Search/Search2DMatrix.swift

@@ -0,0 +1,51 @@
+/**
+ * Question Link: https://leetcode.com/problems/search-a-2d-matrix/
+ * Primary idea: Search col and then binary search row
+ *
+ * Time Complexity: O(log(m + n), Space Complexity: O(1)
+ */
+
+class Search2DMatrix {
+    func searchMatrix(_ matrix: [[Int]], _ target: Int) -> Bool {
+        if matrix.count == 0 || matrix[0].count == 0 {
+            return false
+        }
+        
+        let rowNum = searchRow(matrix, target)
+        return searchCol(matrix[rowNum], target)
+    }
+    
+    private func searchRow(_ matrix: [[Int]], _ target: Int) -> Int {
+        var left = 0, right = matrix.count - 1
+        
+        while left + 1 < right {
+            let mid = (right - left) / 2 + left
+            if matrix[mid][0] == target {
+                return mid
+            } else if matrix[mid][0] < target {
+                left = mid
+            } else {
+                right = mid
+            }
+        }
+        
+        return matrix[right][0] <= target ? right : left
+    }
+    
+    private func searchCol(_ nums: [Int], _ target: Int) -> Bool {
+        var left = 0, right = nums.count - 1
+        
+        while left <= right {
+            let mid = (right - left) / 2 + left
+            if nums[mid] == target {
+                return true
+            } else if nums[mid] < target {
+                left = mid + 1
+            } else {
+                right = mid - 1
+            }
+        }
+        
+        return false
+    }
+}

Diff for: Search/Search2DMatrixII.swift

@@ -0,0 +1,28 @@
+/**
+ * Question Link: https://leetcode.com/problems/search-a-2d-matrix-ii/
+ * Primary idea: Start from last element at first row, then move downwards or backwards
+ *
+ * Time Complexity: O(m + n), Space Complexity: O(1)
+ */
+
+class Search2DMatrixII {
+    func searchMatrix(_ matrix: [[Int]], _ target: Int) -> Bool {
+        guard matrix.count > 0 else {
+            return false
+        }
+    
+        var row = 0, col = matrix[0].count - 1
+        
+        while row < matrix.count && col >= 0 {
+            if matrix[row][col] == target {
+                return true
+            } else if matrix[row][col] < target {
+                row += 1
+            } else {
+                col -= 1
+            }
+        }
+        
+        return false
+    }
+}