[Swift]LeetCode910. 最小差值 II | Smallest Range II

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➤微信公衆號：山青詠芝（shanqingyongzhi）
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Given an array A of integers, for each integer A[i] we need to choose either x = -K or x = K, and add x to A[i] (only once).

After this process, we have some array B.

Return the smallest possible difference between the maximum value of B and the minimum value of B. 

Example 1:

Input: A = [1], K = 0 Output: 0 Explanation: B = [1] 

Example 2:

Input: A = [0,10], K = 2 Output: 6 Explanation: B = [2,8] 

Example 3:

Input: A = [1,3,6], K = 3 Output: 3 Explanation: B = [4,6,3] 

Note:

1. 1 <= A.length <= 10000
2. 0 <= A[i] <= 10000
3. 0 <= K <= 10000

---

給定一個整數數組 A，對於每一個整數 A[i]，咱們能夠選擇 x = -K 或是 x = K，並將 x 加到 A[i] 中。

在此過程以後，咱們獲得一些數組 B。

返回 B 的最大值和 B 的最小值之間可能存在的最小差值。 

示例 1：

輸入：A = [1], K = 0
輸出：0
解釋：B = [1]

示例 2：

輸入：A = [0,10], K = 2
輸出：6
解釋：B = [2,8]

示例 3：

輸入：A = [1,3,6], K = 3
輸出：3
解釋：B = [4,6,3] 

提示：

1. 1 <= A.length <= 10000
2. 0 <= A[i] <= 10000
3. 0 <= K <= 10000

---

Runtime: 152 ms

Memory Usage: 19.2 MB

 1 class Solution {
 2     func smallestRangeII(_ A: [Int], _ K: Int) -> Int {
 3         if A.count <= 1 {
 4             return 0
 5         }
 6         let arr = A.sorted { (num1, num2) -> Bool in
 7             return num1 < num2
 8         }
 9         let min1 = arr.first! + K
10         let max1 = arr.last! - K
11         var res = arr.last! - arr.first!
12         for i in 1 ..< arr.count {
13             let min2 = min(min1, arr[i] - K)
14             let max2 = max(max1, arr[i - 1] + K)
15             let temp = max2 - min2
16             res = temp > res ? res : temp
17         }
18         return res
19     }
20 }

---

164ms

 1 class Solution {
 2     func smallestRangeII(_ A: [Int], _ K: Int) -> Int {
 3         var A = A.sorted()
 4         if A.count == 0 || A.count == 1 {
 5             return 0
 6         }
 7         if A[A.count - 1] - A[0] <= K {
 8             return A[A.count - 1] - A[0]
 9         }
10         var diff = A[A.count - 1] - A[0]
11         for i in 0..<(A.count - 1){
12             let high = max(A[i] + 2 * K, A.last!)
13             let low = min(A[i + 1], A.first! + 2 * K)
14             diff = min(diff, high - low)
15         }
16         return diff
17     }
18 }

---

344ms

 1 class Solution {
 2     func smallestRangeII(_ A: [Int], _ K: Int) -> Int {
 3         var A = A.sorted()
 4         var n:Int = A.count
 5         var mx:Int = A[n - 1]
 6         var mn:Int = A[0]
 7         var res:Int = mx - mn
 8         for i in 0..<(n - 1)
 9         {
10             mx = max(mx, A[i] + 2 * K)
11             mn = min(A[i + 1], A[0] + 2 * K)
12             res = min(res, mx - mn)
13         }
14         return res
15     }
16 }

---

392ms

 1 class Solution {
 2     func smallestRangeII(_ A: [Int], _ K: Int) -> Int {
 3         var newA = Array(Set(A)).sorted()
 4         let n = newA.count - 1
 5         var min = newA[n] - newA[0]
 6         if newA.count <= 1{
 7             return 0
 8         }
 9         for i in 1..<newA.count{
10             var arr = [(newA[0]+K),(newA[i-1] + K),(newA[i] - K),(newA[n] - K)]
11             arr.sort()
12             let temp = arr.last! - arr.first!
13             if temp < min {
14                 min = temp
15             }
16         }
17         return min
18     }
19 }