[Swift]LeetCode164. 最大间距 | Maximum Gap

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：https://www.cnblogs.com/strengthen/p/10126992.html 
➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

热烈欢迎，请直接点击！！！

进入博主App Store主页，下载使用各个作品！！！

注：博主将坚持每月上线一个新app！！！

Given an unsorted array, find the maximum difference between the successive elements in its sorted form.

Return 0 if the array contains less than 2 elements.

Example 1:

Input: [3,6,9,1]
Output: 3
Explanation: The sorted form of the array is [1,3,6,9], either
             (3,6) or (6,9) has the maximum difference 3.

Example 2:

Input: [10]
Output: 0
Explanation: The array contains less than 2 elements, therefore return 0.

Note:

• You may assume all elements in the array are non-negative integers and fit in the 32-bit signed integer range.
• Try to solve it in linear time/space.

---

给定一个无序的数组，找出数组在排序之后，相邻元素之间最大的差值。

如果数组元素个数小于 2，则返回 0。

示例 1:

输入: [3,6,9,1]
输出: 3
解释: 排序后的数组是 [1,3,6,9], 其中相邻元素 (3,6) 和 (6,9) 之间都存在最大差值 3。

示例 2:

输入: [10]
输出: 0
解释: 数组元素个数小于 2，因此返回 0。

说明:

• 你可以假设数组中所有元素都是非负整数，且数值在 32 位有符号整数范围内。
• 请尝试在线性时间复杂度和空间复杂度的条件下解决此问题。

---

48ms

class Solution {
    func maximumGap(_ nums: [Int]) -> Int {
if nums.count < 2 {
        return 0
    }
    var minNums = INTPTR_MAX
    var maxNums = 0
    for i in nums {
        if i < minNums {
            minNums = i
        }
        if i > maxNums {
            maxNums = i
        }
    }
    let gap = max(1,(maxNums - minNums) / (nums.count - 1))
    let bucketNum = (maxNums - minNums) / gap + 1
    var bucketMax = [Int](repeating: 0, count: bucketNum)
    var bucketMin = [Int](repeating: INTPTR_MAX, count: bucketNum)
    var bucketUsed = [Bool](repeating: false, count: bucketNum)
    for i in nums {
        let n = (i - minNums) / gap
        bucketUsed[n] = true
        if i > bucketMax[n] {
            bucketMax[n] = i
        }
        if i < bucketMin[n] {
            bucketMin[n] = i
        }
    }
    var maxGap = 0
    var previousMax = minNums
    for n in (0..<bucketNum) {
        if bucketUsed[n] {
            maxGap = max(maxGap, bucketMin[n] - previousMax)
            previousMax = bucketMax[n]
        }
    }
    return maxGap
    }
}

---

52ms

class Solution {
    func maximumGap(_ nums: [Int]) -> Int {
        var maxGap = 0
        if nums.count < 2 { return maxGap }
        
        var start = 0
        let sortedArray = nums.sorted()
        while (start < sortedArray.count - 1) {
            let diff = sortedArray[start + 1] - sortedArray[start]
            maxGap = max(diff, maxGap)
            start += 1
        }
        
        return maxGap
    }
}

---

56ms

class Solution {
    func maximumGap(_ nums: [Int]) -> Int {
        if nums.isEmpty || nums.count == 1 {return 0}
        var nums: [Int] = nums.sorted(by:<)
        var ret:Int = 0
        for i in 1..<nums.count
        {
            ret = max(ret, nums[i]-nums[i-1])
        }
        return ret
    }
}

---

56ms

class Solution {
    func maximumGap(_ nums: [Int]) -> Int {
        if nums.count < 2 {
            return 0
        }
        var res = 0
        var maxn = 0
        var minn = Int.max
        for i in 0..<nums.count {
            maxn = max(maxn, nums[i])
            minn = min(minn, nums[i])
        }
        
        let n = (maxn - minn) / nums.count + 1
        let buckets = (maxn - minn) / n + 1
        var maxs = Array(repeating: minn-1, count: buckets)
        var mins = Array(repeating: maxn+1, count: buckets)
        for i in 0..<nums.count {
            let index = (nums[i] - minn) / n
            mins[index] = min(mins[index], nums[i])
            maxs[index] = max(maxs[index], nums[i])
        }
        
        var i = 0
        while i < buckets  {
            while maxs[i] == minn - 1 {
                i+=1
            }
            var j = i+1
            if j>=buckets {
                break
            }
            while mins[j] == maxn + 1 {
                j += 1
            }
            res = max(res,mins[j] - maxs[i])
            
            i = j
        }
        
        return res
    }
}