feat: add swift implementation to lcci problem: No.17.04 (#2772)

---------

Co-authored-by: Libin YANG <contact@yanglibin.info>

Author: klever34

lcci/17.04.Missing Number/README.md

@@ -119,6 +119,20 @@ var missingNumber = function (nums) {
 };
 ```
 
+```swift
+class Solution {
+    func missingNumber(_ nums: [Int]) -> Int {
+        let nums = nums.sorted()
+        for (i, x) in nums.enumerated() {
+            if i != x {
+                return i
+            }
+        }
+        return nums.count
+    }
+}
+```
+
 <!-- tabs:end -->
 
 ### 方法二：求和
@@ -206,6 +220,15 @@ var missingNumber = function (nums) {
 };
 ```
 
+```swift
+class Solution {
+    func missingNumber(_ nums: [Int]) -> Int {
+        let n = nums.count
+        return n * (n + 1) / 2 - nums.reduce(0, +)
+    }
+}
+```
+
 <!-- tabs:end -->
 
 ### 方法三：位运算

lcci/17.04.Missing Number/README_EN.md

@@ -119,6 +119,20 @@ var missingNumber = function (nums) {
 };
 ```
 
+```swift
+class Solution {
+    func missingNumber(_ nums: [Int]) -> Int {
+        let nums = nums.sorted()
+        for (i, x) in nums.enumerated() {
+            if i != x {
+                return i
+            }
+        }
+        return nums.count
+    }
+}
+```
+
 <!-- tabs:end -->
 
 ### Solution 2
@@ -202,6 +216,15 @@ var missingNumber = function (nums) {
 };
 ```
 
+```swift
+class Solution {
+    func missingNumber(_ nums: [Int]) -> Int {
+        let n = nums.count
+        return n * (n + 1) / 2 - nums.reduce(0, +)
+    }
+}
+```
+
 <!-- tabs:end -->
 
 ### Solution 3

lcci/17.04.Missing Number/Solution.swift

@@ -0,0 +1,11 @@
+class Solution {
+    func missingNumber(_ nums: [Int]) -> Int {
+        let nums = nums.sorted()
+        for (i, x) in nums.enumerated() {
+            if i != x {
+                return i
+            }
+        }
+        return nums.count
+    }
+}

lcci/17.04.Missing Number/Solution2.swift

@@ -0,0 +1,6 @@
+class Solution {
+    func missingNumber(_ nums: [Int]) -> Int {
+        let n = nums.count
+        return n * (n + 1) / 2 - nums.reduce(0, +)
+    }
+}