13. Roman to Integer

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Given a roman numeral, convert it to an integer.

Input is guaranteed to be within the range from 1 to 3999.

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JAVA:

public class Solution {    public int romanToInt(String s) {        if(s == null || s.length() == 0) return 0;        int[] num = new int[s.length()];        for(int i = 0; i < s.length(); i++) {            switch(s.charAt(i)) {                case 'M':                    num[i] = 1000;                    break;                case 'D':                    num[i] = 500;                    break;                case 'C':                    num[i] = 100;                    break;                case 'L':                    num[i] = 50;                    break;                case 'X':                    num[i] = 10;                    break;                case 'V':                    num[i] = 5;                    break;                case 'I':                    num[i] = 1;                    break;                default:                    return 0;                }            }        int sum = 0;        for(int i = 0; i < num.length-1; i++) {            if (num[i] < num[i+1]) sum -= num[i];            else sum += num[i];        }        sum += num[num.length-1];        return sum;    }}

GO:
go的思路更简洁，使用递归，首先找到给定数字中最大的一个，然后该值减去左边的值再加上右边的值就是得到的结果。

import (    "bytes")var order = [7]byte{'M', 'D', 'C', 'L', 'X', 'V', 'I'}var value = [7]int{1000, 500, 100, 50, 10, 5, 1}func romanToInt(s string) int {    if "" == s {        return 0    }    return convert([]byte(s))}func convert(num []byte) int {    if nil == num || len(num) == 0 {        return 0    }    var idx int    for i, c := range order {        idx = bytes.IndexByte(num, c)        if idx == -1 {            continue        }        if idx > 0 {            return value[i] - convert(num[:idx]) + convert(num[idx+1:])        }        return value[i] + convert(num[idx+1:])    }    return 0}

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