[Swift]LeetCode1054.距离相等的条形码 | Distant Barcodes

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In a warehouse, there is a row of barcodes, where the i-th barcode is barcodes[i].

Rearrange the barcodes so that no two adjacent barcodes are equal.  You may return any answer, and it is guaranteed an answer exists.

Example 1:

Input: [1,1,1,2,2,2]
Output: [2,1,2,1,2,1]

Example 2:

Input: [1,1,1,1,2,2,3,3]
Output: [1,3,1,3,2,1,2,1]

Note:

1. 1 <= barcodes.length <= 10000
2. 1 <= barcodes[i] <= 10000

---

在一个仓库里，有一排条形码，其中第 i 个条形码为 barcodes[i]。

请你重新排列这些条形码，使其中两个相邻的条形码 不能 相等。 你可以返回任何满足该要求的答案，此题保证存在答案。

示例 1：

输入：[1,1,1,2,2,2]
输出：[2,1,2,1,2,1]

示例 2：

输入：[1,1,1,1,2,2,3,3]
输出：[1,3,1,3,2,1,2,1]

提示：

1. 1 <= barcodes.length <= 10000
2. 1 <= barcodes[i] <= 10000

---

Runtime: 476 ms

Memory Usage: 22.2 MB

class Solution {
    func rearrangeBarcodes(_ barcodes: [Int]) -> [Int] {
        let n:Int = barcodes.count
        var cnt:[Int] = [Int](repeating: 0, count: 10005)
        var ret:[Int] = [Int](repeating: 0, count: n)
        for i in 0..<n
        {
            cnt[barcodes[i]] += 1
        }
        var mx:Int = 0
        var id:Int = 0
        for i in 1...10000
        {
            if cnt[i]>mx
            {
                mx = cnt[i]
                id = i
            }
        }
        var j:Int = 0
        while(j < n)
        {
            if cnt[id] > 0
            {
                cnt[id] -= 1
                ret[j] = id
            }
            else
            {
                break
            }
            j += 2
        }
        if j >= n {j = 1}
        for i in 1...10000
        {
            while(cnt[i] > 0)
            {
                cnt[i] -= 1
                ret[j]=i
                j+=2
                if j >= n
                {
                    j = 1
                }
            }
        }
        return ret
    }
}

---

672ms

class Solution {
    func rearrangeBarcodes(_ barcodes: [Int]) -> [Int] {
                
        var map = [Int: Int]()
        for code in barcodes {
            map[code] = map[code, default: 0] + 1
        }
        let sortKeys = map.keys.sorted { map[$0]! > map[$1]! }
        var i = 0, ans = barcodes
        for k in sortKeys {
            for j in 0..<map[k]! {
                ans[i] = k
                i += 2
                if i >= barcodes.count { i = 1 }
            }
        }
        return ans
    }
}

---

700ms

class Solution {
    func rearrangeBarcodes(_ barcodes: [Int]) -> [Int] {
        guard barcodes.count > 1 else { return barcodes }
        
        var counts = [Int: Int]()
        
        for barcode in barcodes {
            counts[barcode] = counts[barcode, default: 0] + 1
        }
        
        let unique = counts.keys.sorted { (k1, k2) -> Bool in
            return counts[k1]! > counts[k2]!
        }
        
        var results = [[Int]]()
        
        let most = unique[0]
        
        for _ in 0..<counts[most]! {
            results.append([most])
        }
        
        var p = 0
        for i in 1..<unique.count {
            let key = unique[i]
            let c = counts[key]!
            
            for _ in 0..<c {
                results[p].append(key)
                p = (p + 1) % results.count
            }
        }
        
        var result = [Int]()
        
        for a in results {
            result += a
        }
        
        return result
    }
}

---

708ms

class Solution {
    func rearrangeBarcodes(_ barcodes: [Int]) -> [Int] {
        var res = Array(repeating: 0, count: barcodes.count)
        var map: [Int: Int] = [:]
        for b in barcodes {
            map[b] = (map[b] ?? 0) + 1
        }
        
        var sortd = map.keys.sorted {
            map[$0]! >  map[$1]!
        }
        
        print(sortd)
        var i = 0
        var sortdInd = 0
        while i < res.count {
            res[i] = sortd[sortdInd];
            if map[sortd[sortdInd]]! - 1 == 0 { 
                sortdInd += 1 
            } else {
                map[sortd[sortdInd]] = map[sortd[sortdInd]]! - 1
            }
            i += 2
            if i >= res.count && i % 2 == 0 {
                i = 1
            }    
        }
        return res
    }
}