如何根據重復值計數對整數陣列進行排序。這里應該首先減少重復次數。
input [5, 2, 1, 2, 4, 4, 1, 1, 2, 3, 3, 6]
OutPut [5, 6, 4, 4, 3, 3, 2, 2, 2, 1, 1, 1]
uj5u.com熱心網友回復：
使用這樣的函式實際上很容易做到.sort/.sorted這一點：
let output = input.sorted { i1, i2 in
i1 != i2
}
uj5u.com熱心網友回復：
使用 Martin 的評論，這是另一種方法，旨在通過使用 swift 提供的一些函式來減少我們自己撰寫的回圈和條件的數量。
// Input
let numbers = [1, 1, 1, 2, 2, 2, 3, 3, 4, 4, 5, 6]
// Count the occurrences based on Martin's comment
let countDict = numbers.reduce(into: [:], { $0[$1, default: 0] = 1 } )
// Get a sorted array of the countDict keys, sorted by value which
// is the number of occurrences
let sortedKeys
= countDict.keys
.sorted { countDict[$0, default: 0] < countDict[$1, default: 0] }
// Initialize an empty array to hold the final sorted numbers
var sortedNumbers: [Int] = []
// Add the elements into the sortedNumbers with in their desired order
for key in sortedKeys {
sortedNumbers.append(contentsOf: repeatElement(key,
count: countDict[key, default: 0]))
}
// prints [5, 6, 4, 4, 3, 3, 1, 1, 1, 2, 2, 2] based on the above input
print(sortedNumbers)
uj5u.com熱心網友回復：
let numbers = [5,2,1,2,4,4,1,1,2,3,3,6]
let sortedNumber = numbers.sorted()
print("Input: ",sortedNumber)
var dict = [Int: Int]()
for item in sortedNumber {
let isExist = dict.contains(where: {$0.key == item})
if !isExist {
dict[item] = 1
} else {
if let value = dict[item] {
dict[item] = value 1
}
}
}
var finalArray = [Int]()
let sortedArray = dict.sorted { (first, second) -> Bool in
return first.value < second.value
}
for d in sortedArray {
for _ in 1...d.value {
finalArray.append(d.key)
}
}
print("Output: ",finalArray)
Input: [1, 1, 1, 2, 2, 2, 3, 3, 4, 4, 5, 6]
Output: [5, 6, 4, 4, 3, 3, 2, 2, 2, 1, 1, 1]
轉載請註明出處，本文鏈接：https://www.uj5u.com/qukuanlian/440615.html
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