Add solution for 4. Median of Two Sorted Arrays.

Author: Charlene Jiang

Search/MedianOfTwoSortedArrays.swift

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+/**
+ * Question Link: https://leetcode.com/problems/median-of-two-sorted-arrays/
+ *
+ * Primary idea: For arrays of m and n numbers, nums1 and nums2, where m <= n.
+ *               To find an index of mid1 in nums1, to separate the arrays into left and right parts:
+ *      nums1[0, 1, ..., mid1 - 1] | nums1[mid1, mid1 + 1, ..., m]
+ *      nums2[0, 1, ..., mid2 - 1] | nums2[mid2, mid2 + 1, ..., n]
+ *
+ *      Make sure:
+ *              count of left = count of right
+ *               max of left <= min of right
+ *
+ * Time Complexity: O(log(n + m)), Space Complexity: O(1)
+ *
+ */
+ 
+ class Solution {
+    func findMedianSortedArrays(nums1: [Int], _ nums2: [Int]) -> Double {
+        let m = nums1.count
+        let n = nums2.count
+        
+        if m > n {
+            return findMedianSortedArrays(nums2, nums1)
+        }
+
+        var halfLength: Int = (m + n + 1) >> 1
+        var b = 0, e = m
+        var maxOfLeft = 0
+        var minOfRight = 0
+                
+        while b <= e {
+            let mid1 = (b + e) >> 1
+            let mid2 = halfLength - mid1
+            
+            if mid1 > 0 && mid2 < n && nums1[mid1 - 1] > nums2[mid2] {
+                e = mid1 - 1
+            } else if mid2 > 0 && mid1 < m && nums1[mid1] < nums2[mid2 - 1] {
+                b = mid1 + 1
+            } else {
+                if mid1 == 0 {
+                    maxOfLeft = nums2[mid2 - 1]
+                } else if mid2 == 0 {
+                    maxOfLeft = nums1[mid1 - 1]
+                } else {
+                    maxOfLeft = max(nums1[mid1 - 1], nums2[mid2 - 1])
+                }
+                
+                if (m + n) % 2 == 1 {
+                    return Double(maxOfLeft)
+                }
+                
+                if mid1 == m {
+                    minOfRight = nums2[mid2]
+                } else if mid2 == n {
+                    minOfRight = nums1[mid1]
+                } else {
+                    minOfRight = min(nums1[mid1], nums2[mid2])
+                }
+                
+                break
+            }
+        }
+        return Double(maxOfLeft + minOfRight) / 2.0
+    }
+}