Swift按照数组元素出现的次数及大小排序

1、已知一个数组，按照单个元素在数组中出现的次数作为重新排序的依据，个数多的排在前面
2、相同个数时候，元素值大的排前面

[1, 2, 2, 3, 5, 5]
经过计算得到的结果是：
[5, 5, 2, 2, 3, 1]

import UIKit

extension Sequence where Iterator.Element == Int{
private func removeRepeats()->[Int]{
let set = Set(self)
return Array(set).sorted {$0>$1}
}

private func countFor(value:Int)->Int{
return filter {$0 == value}.count
}

func sortByRepeatCount()->[Iterator.Element]{
var wets = [[Int]]()
let clearedAry = removeRepeats()
for i in clearedAry{
wets.append([i,countFor(value: i)])
}

wets = wets.sorted {
$0[1] > $1[1]
}

var result = [Int]()
for x in wets{
let i = x[0]
let count = x[1]
for _ in 0..<count{
result.append(i)
}
}

return result
}
}

var ary = [1,1,2,1,3,3,4,5,4,6,6,6]
print(ary.sortByRepeatCount())
//输出 "[6, 6, 6, 1, 1, 1, 4, 4, 3, 3, 5, 2]\n"

extension SequenceType where Generator.Element : Hashable {
func frequencies() -> [Generator.Element:Int] {
var results : [Generator.Element:Int] = [:]
for element in self {
results[element] = (results[element] ?? 0) + 1
}
return results
}
}

let alpha = [2,8,2,6,1,8,2,6,6]
let beta = [6,6,6,2,2,2,8,8,1]

let sorted = alpha.frequencies().sort {
if $0.1 > $1.1 { // if the frequency is higher, return true
return true
} else if $0.1 == $1.1 { // if the frequency is equal
return $0.0 > $1.0 // return value is higher
} else {
return false // else return false
}
}