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

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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

---

给定一个整数数组 A，对于每个整数 A[i]，我们可以选择 x = -K 或是 x = K，并将 x 加到 A[i] 中。

在此过程之后，我们得到一些数组 B。

返回 B 的最大值和 B 的最小值之间可能存在的最小差值。 

示例 1：

输入：A = [1], K = 0
输出：0
解释：B = [1]

示例 2：

输入：A = [0,10], K = 2
输出：6
解释：B = [2,8]

示例 3：

输入：A = [1,3,6], K = 3
输出：3
解释：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

class Solution {
    func smallestRangeII(_ A: [Int], _ K: Int) -> Int {
        if A.count <= 1 {
            return 0
        }
        let arr = A.sorted { (num1, num2) -> Bool in
            return num1 < num2
        }
        let min1 = arr.first! + K
        let max1 = arr.last! - K
        var res = arr.last! - arr.first!
        for i in 1 ..< arr.count {
            let min2 = min(min1, arr[i] - K)
            let max2 = max(max1, arr[i - 1] + K)
            let temp = max2 - min2
            res = temp > res ? res : temp
        }
        return res
    }
}

---

164ms

class Solution {
    func smallestRangeII(_ A: [Int], _ K: Int) -> Int {
        var A = A.sorted()
        if A.count == 0 || A.count == 1 {
            return 0
        }
        if A[A.count - 1] - A[0] <= K {
            return A[A.count - 1] - A[0]
        }
        var diff = A[A.count - 1] - A[0]
        for i in 0..<(A.count - 1){
            let high = max(A[i] + 2 * K, A.last!)
            let low = min(A[i + 1], A.first! + 2 * K)
            diff = min(diff, high - low)
        }
        return diff
    }
}

---

344ms

class Solution {
    func smallestRangeII(_ A: [Int], _ K: Int) -> Int {
        var A = A.sorted()
        var n:Int = A.count
        var mx:Int = A[n - 1]
        var mn:Int = A[0]
        var res:Int = mx - mn
        for i in 0..<(n - 1)
        {
            mx = max(mx, A[i] + 2 * K)
            mn = min(A[i + 1], A[0] + 2 * K)
            res = min(res, mx - mn)
        }
        return res
    }
}

---

392ms

class Solution {
    func smallestRangeII(_ A: [Int], _ K: Int) -> Int {
        var newA = Array(Set(A)).sorted()
        let n = newA.count - 1
        var min = newA[n] - newA[0]
        if newA.count <= 1{
            return 0
        }
        for i in 1..<newA.count{
            var arr = [(newA[0]+K),(newA[i-1] + K),(newA[i] - K),(newA[n] - K)]
            arr.sort()
            let temp = arr.last! - arr.first!
            if temp < min {
                min = temp
            }
        }
        return min
    }
}