Question

我有数组，需要在没有Array.reverse方法的情况下将其反转，只需使用for循环。

var names:[String] = ["Apple", "Microsoft", "Sony", "Lenovo", "Asus"]

Answer 1

这是@Abhinav的回答转换为 Swift 2.2 ：

var names: [String] = ["Apple", "Microsoft", "Sony", "Lenovo", "Asus"]

var reversedNames = [String]()

for arrayIndex in (names.count - 1).stride(through: 0, by: -1) {
    reversedNames.append(names[arrayIndex])
}

使用此代码不应该为您提供有关C风格for循环弃用或--使用的任何错误或警告。

Swift 3 - 当前：

let names: [String] = ["Apple", "Microsoft", "Sony", "Lenovo", "Asus"]

var reversedNames = [String]()

for arrayIndex in stride(from: names.count - 1, through: 0, by: -1) {
    reversedNames.append(names[arrayIndex])
}

或者，您可以正常循环并每次减去：

let names = ["Apple", "Microsoft", "Sony", "Lenovo", "Asus"]

let totalIndices = names.count - 1 // We get this value one time instead of once per iteration.

var reversedNames = [String]()

for arrayIndex in 0...totalIndices {
    reversedNames.append(names[totalIndices - arrayIndex])
}

Answer 2

Swift 3:

var names:[String] = ["Apple", "Microsoft", "Sony", "Lenovo", "Asus"]

var reversedNames : [String] = Array(names.reversed())

print(reversedNames)  // ["Asus", "Lenovo", "Sony", "Microsoft", "Apple"]

Answer 3

¿loop是必须的吗？如果没有，您可以使用reduce。

我猜这是没有 reverse（）方法实现它的最短路径（ Swift 3.0.1 ）：

["Apple", "Microsoft", "Sony", "Lenovo", "Asus"].reduce([],{ [$1] + $0 })

Answer 4

var names:[String] = [ "A", "B", "C", "D", "E","F","G"]
var c = names.count - 1
var i = 0
while i < c {
    swap(&names[i++],&names[c--])
}

Answer 5

你走了：

var names:[String] = ["Apple", "Microsoft", "Sony", "Lenovo", "Asus"]

var reversedNames = [String]()

for var arrayIndex = names.count - 1 ; arrayIndex >= 0 ; arrayIndex-- {
    reversedNames.append(names[arrayIndex])
}

Answer 6

还有stride生成反向索引：

let names = ["Apple", "Microsoft", "Sony", "Lenovo", "Asus"]

var reversed = [String]()

for index in (names.count - 1).stride(to: -1, by: -1) {
    reversed.append(names[index])
}

它也适用于map：

let reversed = (names.count - 1).stride(to: -1, by: -1).map { names[$0] }

注意：stride的索引为1，而不是0，与其他Swift序列相反。

但是，对于将来阅读此内容的任何人：使用.reverse() 代替实际反转数组，这是预期的方式。

Answer 7

这里是swift 3的代码

let array = ["IOS A", "IOS B", "IOS C"]
    for item in array.reversed() {
    print("Found \(item)")
    }

Answer 8

只需要make（names.count / 2）通过数组。在进行交换时，无需声明临时变量......它是隐含的。

var names:[String] = ["Apple", "Microsoft", "Sony", "Lenovo", "Asus"]
let count = names.count
for i in 0..<count/2 {
   (names[i],names[count - i - 1])  = (names[count - i - 1],names[i])
}
// Yields: ["Asus", "Lenovo", "Sony", "Microsoft", "Apple"]

Answer 9

像这样，也许：

names = names.enumerate().map() { ($0.index, $0.element) }.sort() { $0.0 > $1.0 }.map() { $0.1 }

哦，等等..我有使用for循环，对吗？那么可能就是这样：

for (index, name) in names.enumerate().map({($0.index, $0.element)}).sort({$0.0 > $1.0}).map({$0.1}).enumerate() {
    names[index] = name
}

Answer 10

忽略对空虚的检查..

var names:[String] = ["Apple", "Microsoft", "Sony", "Lenovo", "Asus"]

var reversedNames: [String]
reversedNames = []

for name in names {
    reversedNames.insert((name), atIndex:0)
}

Answer 11

你是说

吗？

var names = ["Apple", "Microsoft", "Sony", "Lenovo", "Asus"]
var newNames = [String]()

for var i = names.count - 1; i >= 0 ; i-- {
  newNames.append(names[i])
}
names = newNames

或

names.map {newNames.insert($0, atIndex: 0)}
names = newNames

Answer 12

我是这样做的，并没有对Swift 3的警告

let names = ["Apple", "Microsoft", "Sony", "Lenovo", "Asus"]
var reversedNames = [String]()

for name in names.enumerate() {
  let newIndex = names.count - 1 - name.index
  reversedNames.append(names[newIndex])
}

或只是简单地

reversedNames = names.reverse()

Answer 13

这适用于任何大小的数组。

import Cocoa

var names:[String] = [ "A", "B", "C", "D", "E","F"]
var c = names.count - 1
for i in 0...(c/2-1) { swap(&names[i],&names[c-i]) }

print(names)

Answer 14

func reverse(array: inout [String]) {
    if array.isEmpty { return }
    var f = array.startIndex
    var l = array.index(before: array.endIndex)
    while f < l {
        swap(array: &array, f, l)
        array.formIndex(after: &f)
        array.formIndex(before: &l)
    }
}

private func swap( array: inout [String], _ i: Int, _ j: Int) {
    guard i != j else { return }
    let tmp = array[i]
    array[i] = array[j]
    array[j] = tmp
}

或者你可以写课程的扩展

Answer 15

var rArray : [Int] = [2,5,6,8,3,8,9,10]
var ReArray = [Int]()
var a : Int = 1

func reversed (_ array: [Int]) -> [Int] {
    for i in array {
        ReArray.append(array[array.count-a])
        a += 1
    }

    rArray = ReArray

    return rArray
}

reversed(rArray)

print(rArray)

Answer 16

您可以使用swift3文档：

let names = ["Chris", "Alex", "Ewa", "Barry", "Daniella"]
let reversedNames = names.sorted(by: >)

// reversedNames is equal to:
//   ["Ewa", "Daniella", "Chris", "Barry", "Alex"]

Answer 17

执行此操作以进行反向排序。

 let unsortedArray: [String] = ["X", "B", "M", "P", "Z"]
        // reverse sorting
        let reversedArray = unsortedArray.sorted() {$0 > $1}

        print(reversedArray) // ["Z", "X", "P", "M", "B"]

Answer 18

这是最简单的方法。

let names:[String] = ["Apple", "Microsoft", "Sony", "Lenovo", "Asus"]

var reversedNames = [String]()

for name in names {
    reversedNames.insert(name, at: 0)
}

Answer 19

x_names = ['John', 'Jill', 'Sandy']

y_ages = 8,11,10

Trace00 = dict(type = 'bar', x=x_names, y= y_ages)

plotly.offline.iplot([Trace00])

Answer 20

var arr = [1, 2, 3, 4, 5]   // Array we want to reverse
var reverse: [Int]!      // Array where we store reversed values

reverse = arr

for i in 0...(arr.count - 1) {

    reverse[i] = arr.last!  // Adding last value of original array at reverse[0]
    arr.removeLast()        // removing last value & repeating above step.
}

print("Reverse : \(reverse!)")

更简单的方法：）

Answer 21

我喜欢简单的代码。

变量名：[String] = [“ Apple”，“ Microsoft”，“ Sony”，“ Lenovo”，“ Asus”]

var reversedNames = [""]

for name in names {
    reversedNames.insert(name, at: 0)
}

print(reversedNames)

Answer 22

Swift 5：

let names: [String] = ["Apple", "Microsoft", "Sony", "Lenovo", "Asus"]

var reversenames: [String] = []

让计数 = names.count

用于索引在 0..

reversenames.insert(names[count-index-1], at: index)

}

打印（反名）

如何在不使用＆＃34; .reverse（）＆＃34;的情况下在Swift中反转数组？

22 个答案: