[LeetCode 013] Roman to Integer

Roman to Integer

  • 从后往前遍历字符串
  • 当前一个字符代表的数值比后一个小时,用最终结果减去这个数值
  • 否则,用最终结果加上这个数值

Implementation

public class Solution {
    public int romanToInt(String s) {
        if (s == null || s.length() == 0)
            return 0;
        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 length = s.length();
	    int result = map.get(s.charAt(length - 1));
	    int current = result;
	    for (int i = length - 2; i >= 0; i--) {
	        int pre = map.get(s.charAt(i));
	        result += (pre < current? -pre: pre);
	        current = pre;
	    }
	    return result;
    }
}
原文地址:https://www.cnblogs.com/Victor-Han/p/5195000.html