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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.A

LeetCode 401. 二进制手表 scala解法

object Solution { def readBinaryWatch(turnedOn: Int): List[String] = { var ans = scala.collection.mutable.ArrayBuffer[String]() for(h

[leetcode]13. 罗马数字转整数

class Solution { public int romanToInt(String s) { int ans = 0; for(int i = 0;i < s.length();i++){ if(i == s.length()-1