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.