leetcode--Roman to Integer-优快云博客

本文介绍了一种将罗马数字转换为整数的算法实现，采用贪婪算法思想，通过将字符串切分为尽可能长的子串来完成转换。适用于1到3999范围内的罗马数字。

Given a roman numeral, convert it to an integer.

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

public class Solution {
   /**
	 * The algorithm is kind of greedy algorithm: cut the string into substrings such that
	 * the length of each substring is as large as possible.<br> 
	 * @param s -string, Roman expression of an integer
	 * @return int
	 * @author Averill Zheng
	 * @version 2014-06-22
	 * @since JDk 1.7
	 */
	public int romanToInt(String s) {
		int sum = 0;
		Map<String, Integer> roman2Int = new HashMap<String, Integer>();
		int[] units = new int[]{1000,900,500,400,100,90,50,40,10,9,5,4,1};
        String[] romanNumeral = new String[]{"M","CM", "D", "CD", "C", "XC", "L",
        		"XL", "X", "IX", "V", "IV", "I"};
		for(int i = 0; i < 13; ++i)
			roman2Int.put(romanNumeral[i], units[i]);
		int length = s.length();
		int i = 0;
		while(i < length){
			if(i + 1 < length && (s.substring(i, i + 2).equals("CM") || s.substring(i, i + 2).equals("CD") ||
					s.substring(i, i + 2).equals("XC") || s.substring(i, i + 2).equals("XL") ||
					s.substring(i, i + 2).equals("IX") || s.substring(i, i + 2).equals("IV"))){
				sum += roman2Int.get(s.substring(i, i + 2));
				i += 2;
			}			
			else{
				sum += roman2Int.get(s.substring(i, i + 1));
				i += 1;
			}
		}
		return sum; 
    }
}

转载于:https://www.cnblogs.com/averillzheng/p/3803774.html