解题思路参考:
剑指 Offer 38. 字符串的排列(回溯法,清晰图解)
object Solution {
def permutation(s: String): Array[String] = {
var res = new scala.collection.mutable.ArrayBuffer[String]
var chars = s.toCharArray
val loop = new scala.util.control.Breaks
val swap = (a:Int,b:Int) => {
val tmp = chars(a)
chars(a) = chars(b)
chars(b) = tmp
}
def dfs(x:Int) {
if(x == chars.length -1){
res += chars.mkString
}
var set = new scala.collection.mutable.HashSet[Char]
for( i <- x until chars.length){
loop.breakable(
if (set.contains(chars(i))){
loop.break
}
else{
set.add(chars(i))
swap(i,x)
dfs(x+1)
swap(i,x)
}
)
}
}
dfs(0)
res.toArray
}
}
Q.E.D.