[LeetCode]13. Roman to Integer羅馬數字轉整數
阿新 • • 發佈:2019-01-03
instead placed man for sym () 兩種 字符 together
, which is simply . There are six instances where subtraction is used:
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
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
I
can be placed beforeV
(5) andX
(10) to make 4 and 9.X
can be placed beforeL
(50) andC
(100) to make 40 and 90.C
can be placed beforeD
(500) andM
(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.
跟上一題相反,要把羅馬數字轉成阿拉伯數字,觀察之後我們發現,可以從左到右遍歷羅馬數字的字符串,無非兩種可能
1.當前字符是最後一位或者表示數字不小於後面一位字符表示數字,類似VII,就是往上加,先是5,然後是5+1,最後6+1
2.當前字符表示數字比後一位字符小,類似IV,就需要減去當前位的數字
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; } }
[LeetCode]13. Roman to Integer羅馬數字轉整數