Swift按照数组元素出现的次数及大小排序

要求如下:

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

例子:

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

一种解决方法是:

import UIKitextension 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"

别的网友提供了更直观更简单的方法:

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    }}

注意后一种解决办法只能在Swift2.x中运行,如果要在Swift3中运行需要略做修改,请参考我写的另一篇博文:

Swift3中如何为Array写一个限定Type的扩展