LeetCode解题报告--Roman to Integer

题目：罗马数字转为阿拉伯数字
Given a roman numeral, convert it to an integer.

Input is guaranteed to be within the range from 1 to 3999.
分析：题意：将给定的罗马数字转为阿拉伯数字
从前往后遍历罗马数字，如果某个数比前一个数小，则把该数加入到结果中；
反之，则在结果中两次减去前一个数并加上当前这个数；

java 代码：（accepted）

import java.util.HashMap;

import sun.print.resources.serviceui;

public class Roman_To_Integer {

    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        String s = "MMCMLXXXIV";
        //String s = "MMMCMXCIX";
        System.out.println("roman to int: " + romanToInt(s));
    }

    public static int romanToInt(String s) {

        //Genericity to set table to search for
        HashMap<Character, Integer> map = new HashMap<Character, Integer>();
        map.put(‘I‘, 1);
        map.put(‘V‘, 5);
        map.put(‘X‘, 10);
        map.put(‘L‘, 50);
        map.put(‘C‘, 100);
        map.put(‘D‘, 500);
        map.put(‘M‘, 1000);

        int num = 0;
        for (int i = 0; i < s.length(); i++) {
            if (i + 1 <= s.length() - 1 && (map.get(s.charAt(i)) < map.get(s.charAt(i + 1)))) {
                num = num + map.get(s.charAt(i + 1)) - map.get(s.charAt(i));
            //  System.out.println("flag1: " + num + "   " + s.charAt(i) + "  " + s.charAt(i + 1));
                i ++;//When the (i + 1)th lager than (i)th i have to plus 1 to skip the (i + 1)th witch have included
            } else {
                num = num + map.get(s.charAt(i));
            //  System.out.println("flag2: " + num + "   "  + s.charAt(i));
            }
        }

        return num;
    }
}

测试结果：

roman to int: 2984

相关代码放在个人github:https://github.com/gannyee/LeetCode/tree/master/src