[Swift]LeetCode908. 最小差值 I | Smallest Range I

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Given an array A of integers, for each integer A[i] we may choose any x with -K <= x <= K, and add x to A[i].

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: 0
Explanation: B = [3,3,3] or B = [4,4,4] 

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]，我们可以选择任意 x 满足 -K <= 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：0
解释：B = [3,3,3] 或 B = [4,4,4] 

提示：

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

---

16ms

class Solution {
    func smallestRangeI(_ A: [Int], _ K: Int) -> Int {
        var smallest = A[0], largest = A[0]
        for num in A {
            if num < smallest { smallest = num }
            else if num > largest { largest = num }
        }
        let diff = largest - smallest
        if diff <= 2 * K {
            return 0
        } else {
            return diff - 2 * K
        }
    }
}

---

120ms

class Solution {
    func smallestRangeI(_ A: [Int], _ K: Int) -> Int {
        var ma = Int.min
        var mi = Int.max
        for num in A {
            ma = max(ma,num)
            mi = min(mi,num)
        }
        return (ma-mi) <= 2*K ? 0 : ma-mi-2*K
    }
}

---

124ms

class Solution {
    func smallestRangeI(_ A: [Int], _ K: Int) -> Int {
        guard A.count > 1 else {
            return 0
        }
        let mx = A.max()!
        let mi = A.min()!
        return max(0, mx - mi - K * 2)
    }
}

---

124ms

class Solution {
    func smallestRangeI(_ A: [Int], _ K: Int) -> Int {
        var A = A.sorted()
        guard let first = A.first, let last = A.last else {
            return 0
        }
        return (last - first - 2 * K) > 0 ? (last - first - 2 * K) : 0
    }
}

---

Runtime: 140 ms

Memory Usage: 19 MB

class Solution {
    //抓住最大值与最大值之间的重点关系，
    //因为其他数可以通过加减K靠近极值
    func smallestRangeI(_ A: [Int], _ K: Int) -> Int {
        //如果数组只有一个数字则返回0
        if A.count==1{return 0}
        //初始化最大值
        var maxNum:Int = A[0]
        //初始化最小值
        var minNum:Int = A[0]
        //遍历数组
        for i in 0..<A.count
        {            
            if A[i] > maxNum
            {
                maxNum = A[i] 
            }
            if A[i] < minNum
            {
                minNum = A[i] 
            }
        }
        if minNum+2*K>=maxNum
        {
            return 0
        }
        else
        {
            return maxNum-minNum-2*K
        }
    }
}

---

148ms

class Solution {
    func smallestRangeI(_ A: [Int], _ K: Int) -> Int {
        return max(0, (A.max() ?? 0) - (A.min() ?? 0) - 2*K)
    }
}

---

180ms

class Solution {
    func smallestRangeI(_ A: [Int], _ K: Int) -> Int {
        guard A.count != 1 else {
            return 0
        }
        var max = A[0]
        var min = A[0]
        for i in A {
            max = i > max ? i : max
            min = i < min ? i : min
        }
        var diff = max - min
        return abs(K * 2) >= diff ? 0 : diff - abs(K * 2)
    }
}