leetcode刷题笔记十三 罗马数字转数字 Scala版本

leetcode刷题笔记十三 罗马数字转数字 Scala版本

源地址：罗马数字转数字

问题描述：

oman numerals are represented by seven different symbols: I, V, X, L, C, D and M.

Symbol       Value
I             1
V             5
X             10
L             50
C             100
D             500
M             1000

For example, two is written as II in Roman numeral, just two one's added together. Twelve is written as, XII, which is simply X + II. The number twenty seven is written as XXVII, which is XX + V + II.

Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:

• I can be placed before V (5) and X (10) to make 4 and 9.
• X can be placed before L (50) and C (100) to make 40 and 90.
• C can be placed before D (500) and M (1000) to make 400 and 900.

Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999.

Example 1:

Input: "III"
Output: 3

Example 2:

Input: "IV"
Output: 4

Example 3:

Input: "IX"
Output: 9

Example 4:

Input: "LVIII"
Output: 58
Explanation: L = 50, V= 5, III = 3.

Example 5:

Input: "MCMXCIV"
Output: 1994
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

简要思路分析：

与数字转罗马字母类似，通过映射或者数组建立罗马字母与数值间的对应关系，通过分析罗马字母的构成特点，若s(i-1) < s(i) ,如IX意味着需要将sum值减去s(i-1),否则加上。为了方便考虑边界问题，我们增加了O->0,置于s串末尾。

代码补充：

object Solution {
 def romanToInt(s: String): Int = {
      val dict : Map[Char,Int] = Map(
        'O' -> 0,
        'I' -> 1,
        'V' -> 5,
        'X' -> 10,
        'L' -> 50,
        'C' -> 100,
        'D' -> 500,
        'M' -> 1000
      )
      var ans:Int = 0
      var sStr = s + 'O'
      for (i <- 0 to s.length-1){
        //println("-------" + i + "--------")
        //println("s(i): " + dict(s(i)))
        //println("s(i+1): " + dict(s(i+1)))
        if (dict(sStr(i)) < dict(sStr(i+1))){
          ans = ans - dict(s(i))
        }
        else
        {ans = ans + dict(sStr(i))}
        //println("ans: " + ans)
      }

      return ans
    }
}