[LeetCode]13. Roman to Integer罗马数字转整数

Posted 2021-01-31 jchen104

Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.

Symbol       Value
I             1
V             5
X             10
L             50
C             100
D             500
M             1000

For example, two is written as II in Roman numeral, just two one‘s added together. Twelve is written as, XII, which is simply X + II. The number twenty seven is written as XXVII, which is XX + V + II.

Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:

• I can be placed before V (5) and X (10) to make 4 and 9. 
• X can be placed before L (50) and C (100) to make 40 and 90. 
• C can be placed before D (500) and M (1000) to make 400 and 900.

Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999.

Example 1:

Input: "III"
Output: 3

Example 2:

Input: "IV"
Output: 4

Example 3:

Input: "IX"
Output: 9

Example 4:

Input: "LVIII"
Output: 58
Explanation: L = 50, V= 5, III = 3.

Example 5:

Input: "MCMXCIV"
Output: 1994
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

跟上一题相反，要把罗马数字转成阿拉伯数字，观察之后我们发现，可以从左到右遍历罗马数字的字符串，无非两种可能
1.当前字符是最后一位或者表示数字不小于后面一位字符表示数字，类似VII，就是往上加，先是5，然后是5+1,最后6+1
2.当前字符表示数字比后一位字符小，类似IV，就需要减去当前位的数字

class Solution {
    public int romanToInt(String s) {
        if (s == null || s.length() == 0)    return -1;
        Map<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 res = 0;
        for(int i=0;i<s.length();i++){
            int val=map.get(s.charAt(i));
            if(i==s.length()-1 || map.get(s.charAt(i+1))<=map.get(s.charAt(i))){
                res=res+val;
            }else{
                res=res-val;
            }
        }
        return res;

    }
}