Question

请原谅我，因为我是这种语言的新手。

编辑：有没有办法扭转数组元素的位置？

我正在尝试创建一个函数来测试给定的输入，如果它是一个回文。我试图避免使用reverse（）

来使用函数

let word = ["T","E","S","T"]
var temp = [String]()
let index_count = 3

for words in word{
    var text:String = words
    print(text)
    temp.insert(text, atIndex:index_count)
    index_count = index_count - 1
}

Answer 1

您的方法可以用于反转数组。但你必须这样做将原始数组的每个元素插入开始位置 目标数组（将其他元素移动到末尾）：

// Swift 2.2:
let word = ["T", "E", "S", "T"]
var reversed = [String]()
for char in word {
    reversed.insert(char, atIndex: 0)
}
print(reversed) // ["T", "S", "E", "T"]

// Swift 3:
let word = ["T", "E", "S", "T"]
var reversed = [String]()
for char in word {
    reversed.insert(char, at: 0)
}
print(reversed) // ["T", "S", "E", "T"]

可以直接对字符串的字符进行相同的操作：

// Swift 2.2:
let word = "TEST"
var reversed = ""
for char in word.characters {
    reversed.insert(char, atIndex: reversed.startIndex)
}
print(reversed) // "TSET"

// Swift 3:
let word = "TEST"
var reversed = ""
for char in word.characters {
    reversed.insert(char, at: reversed.startIndex)
}
print(reversed)

Answer 2

斯威夫特 5

impl<'a, T: 'a> RefIterable<'a> for Vec<T> {
    type Item = T;
    type Iter = std::slice::Iter<'a, T>;
    
    fn refs(&'a self) -> std::slice::Iter<'a, T> {
        self.as_slice().iter()
    }
}

Answer 3

这是我的解决方案：

extension String {
    func customReverse() -> String {
        var chars = Array(self.characters)
        let count = chars.count

        for i in 0 ..< count/2 {
            swap(&chars[i], &chars[count - 1 - i])
        }

        return String(chars)
    }
}

let input = "abcdef"
let output = input.customReverse()
print(output)

您可以尝试here。

Answer 4

 static func reverseString(str : String) {
        var data = Array(str)
        var i = 0// initial
        var j = data.count // final
        //either j or i for while , data.count/2 buz only half we need check
        while j != data.count/2  {
            print("befor i:\(i) j:\(j)" )
            j = j-1 
            data.swapAt(i, j) //  swapAt API avalible only for array in swift
            i = i+1 
        }
        print(String(data))
    }

Answer 5

解决方案1

let word = "aabbaa"

let chars = word.characters
let half = chars.count / 2
let leftSide = Array(chars)[0..<half]
let rightSide = Array(chars.reverse())[0..<half]
let palindrome = leftSide == rightSide

解决方案2

var palindrome = true
let chars = Array(word.characters)
for i in 0 ..< (chars.count / 2) {
    if chars[i] != chars[chars.count - 1 - i] {
        palindrome = false
        break
    }
}
print(palindrome)

Answer 6

另一种逆转解决方案：

var original : String = "Test"
var reversed : String = ""
var c = original.characters
for _ in 0..<c.count{
    reversed.append(c.popLast()!)
}

它只是附加弹出的旧字符串的每个元素，从最后一个元素开始并朝着第一个元素开始

不使用预定义函数的反向字符串

6 个答案:

解决方案1

解决方案2

不使用预定义函数的反向字符串

6 个答案:

解决方案1 ​​

解决方案2

解决方案1