leetcode-13 Roman to Integer

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问题描述：Given a roman numeral, convert it to an integer.

Input is guaranteed to be within the range from 1 to3999.

问题分析：{‘M‘,‘D‘,‘C‘,‘L‘,‘X‘,‘V‘,‘I‘};分别对应{1000, 500, 100, 50, 10, 5, 1};

当出现类似CM情况时，即较小者在较大者前面，则应该用较大者减去较小者，即1000-100=900；

否则则直接相加；

这里用到hashMap进行存储，get()的时间复杂度为O(1),用空间换时间

仅需注意条件判断及进位情况即可

代码：

public class Solution {
	public int romanToInt(String s) {		
		if(s == null || s.length() == 0)
			return 0;
		
		char[] numberchars = {'M','D','C','L','X','V','I'};
		int[] values = {1000, 500, 100, 50, 10, 5, 1};
		
		HashMap<Character, Integer> map = new HashMap<>();
		for(int i = 0; i < numberchars.length; i++)
		{
			map.put(numberchars[i], values[i]);
		}
				
		int result = 0;
		char[] datas = s.toCharArray();
		int i = 1;
		int last = map.get(datas[0]);
		int now = 0;
		
		while(i <= datas.length)
		{
			if( i <= datas.length - 1)
			{
				now = map.get(datas[i]);

				if(now > last)
				{
					result += (now - last);
					++i;
					if(i < datas.length)
						last = map.get(datas[i]);
				}
				else
				{
					result += last;	
					last = now;					
				}
			}
			else
			{
				result += last;
				last = now;
			}		
			++i;
		}
		
		return result;		
	}
}

leetcode-13 Roman to Integer

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原文地址：http://blog.csdn.net/woliuyunyicai/article/details/44984179