对于同时具有String和Int值（每个中的一个）的String，可以进行简单排序，将按数字顺序排序的项目作为主要顺序，按字母顺序排列为次要顺序
var nameArray = ["Dave7", "Bob8", "Cathy9", "Henry10", "Susan10", "Pat11", "Steve12", "Dan12", "Ken1", "Sean2", "Howard3", "Dixie3", "Newman5", "Billy6"]
var sortedNameArray = nameArray.sort { $0.compare($1, options: .NumericSearch) == .OrderedAscending }
print(sortedNameArray) // gives the following:
不想要这个 - ＆gt; ["Billy6", "Bob8", "Cathy9", "Dan12", "Dave7", "Dixie3", "Henry10", "Howard3", "Ken1", "Newman5", "Pat11", "Sean2", "Steve12", "Susan10"]
即使使用.NumericSearch，结果也是按字母顺序排列的。
我能够使用自定义二叉树获得所需的结果。结果如下：
Ken1 Sean2 Dixie3 Howard3 Newman5 Billy6 Dave7 Bob8 Cathy9 Henry10 Susan10 Pat11 Dan12 Steve12
但是有更简单的解决方案吗？
extension String {
var integerValue: Int? {
return Int(self)
}
}
func extractValueFromString(theString:String)->Int{
var catNumber: [Character] = []
//print("theString \(theString)")
for character in theString.characters{
var characterString = String(character)
if var value = characterString.integerValue { //if we don't check program crashes
//if numberSet.contains(Int(String(character))!) { //another way to check but redundant here
catNumber.append(character)
//print(catNumber)
// }
}
}
let numberString = String(catNumber)
return Int(numberString)!
}
class Node{
//nodes now only arrange strings
var data = ""
var value = Int()
var left:Node?;
var right:Node?;
deinit {
//print("deleting \(data)")
// print("node deleted")
}
init(data:String){
self.data = data;
//print(data)
}
}
class binaryTreeSort{
var root:Node?
init(){
}
deinit {
//print("tree deleted")
}
func getRoot()->Node{
return root!
}
func insertNewValue(data:String){
let newNode = Node(data:data)
var node:Node? = root
if (node == nil){
root = newNode
}
while (node != nil) {
let currentValue = node?.data
if currentValue == ""{
node?.data = data
return
}
if currentValue == data {
//we don't want duplicates.
return
}
if extractValueFromString(currentValue!) < extractValueFromString(data) {
if (node!.right != nil) {
node = node!.right
//print("Going Right at data \(node!.data)")
}else{
node!.right = newNode
//print("Going New Right at data \(node!.data)")
return
}
}else if extractValueFromString(currentValue!) == extractValueFromString(data){
if currentValue < data {
if (node!.right != nil) {
node = node!.right
//print("Going Right at data \(node!.data)")
}else{
node!.right = newNode
//print("Going New Right at data \(node!.data)")
return
}
}else{
if (node!.left != nil) {
//print("Going Left at data \(node!.data)")
node = node!.left
}else{
node!.left = newNode
//print("Going New Left at data \(node!.data)")
return
}
}
}
else{
if (node!.left != nil) {
//print("Going Left at data \(node!.data)")
node = node!.left
}else{
node!.left = newNode
//print("Going New Left at data \(node!.data)")
return
}
}
}
}
func inorderPrint(baseNode:Node){
if(baseNode.left != nil)
{
inorderPrint(baseNode.left!);
//print(" \(baseNode.data)")
}
print("\(baseNode.data)")
if(baseNode.right != nil)
{
inorderPrint(baseNode.right!)
//print(" \(baseNode.data)")
}
}
func reverseOrderPrint(baseNode:Node){
if(baseNode.right != nil)
{
reverseOrderPrint(baseNode.right!)
//print(" \(baseNode.data)")
}
print("\(baseNode.data)")
if(baseNode.left != nil)
{
reverseOrderPrint(baseNode.left!);
//print(" \(baseNode.data)")
}
}
}
var myBinaryTreeSort:binaryTreeSort? = binaryTreeSort()
for item in nameArray{
//print(item)
myBinaryTreeSort!.insertNewValue(item)
}
myBinaryTreeSort!.inorderPrint(myBinaryTreeSort!.getRoot())
print("---------------")
myBinaryTreeSort!.reverseOrderPrint(myBinaryTreeSort!.getRoot())
myBinaryTreeSort = nil //delete the tree
答案 0 :(得分：1)
如果您有一个数组，则可以使用自定义闭包进行排序。
例如：
nameArray.sort({extractValueFromString($0) < extractValueFromString($1)})
会让你亲近。您只需检查它们是否相等，然后返回$0 < $1。
答案 1 :(得分：1)
以下是我解决这个问题的方法，做了与@Lou-Franco提到的类似的事情：
func endInteger(word: String) -> Int {
if let range = word.rangeOfCharacterFromSet(NSCharacterSet.decimalDigitCharacterSet()){
let numberSubstring = word.substringFromIndex(range.startIndex)
return Int(numberSubstring) ?? 0
}
return 0
}
let sortedArray = yourArray.sort{endInteger($1) > endInteger($0)}
答案 2 :(得分：1)
使用map将名称拆分为多个部分，按数字和名称排序，然后映射以恢复原始名称：
func splitName(name:String) -> (String, Int) {
if let range = name.rangeOfCharacterFromSet(NSCharacterSet.decimalDigitCharacterSet()) {
return (name[name.startIndex..<range.startIndex], Int(name[range.startIndex..<name.endIndex])!)
} else {
return (name, 0)
}
}
print(nameArray.map(splitName).sort({ lhs, rhs in
if lhs.1 < rhs.1 {
return true
} else if lhs.1 > rhs.1 {
return false
} else {
return lhs.0 < rhs.0
}
}).map({ "\($0.0)\($0.1)" }))
可以做的其他一些方法是将元组的元素0保持为全名（带数字），然后最终的地图变为map({ $0.0 })根据大小，这可能比分裂更优化每次比较时的名称。