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 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: C = 100, L = 50, XXX = 30 and III = 3.
Example 5:
Input: "MCMXCIV" Output: 1994 Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.
code1:
public int romanToInt(String s) { /*int[] nums={1000,900,500,400,100,90,50,40,10,9,5,4,1}; String[] romans={"M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"}; int count=0; int result=0; while(count<s.length()){ String res1=""; if(count<s.length()-1) res1=s.substring(count,count+2); String res2=s.substring(count,count+1); for(int i=0;i<romans.length;i++){ if(romans[i].equals(res1)){ count+=2; result+=nums[i]; break; }else if(romans[i].equals(res2)){ count++; result+=nums[i]; break; } } } return result; }
code2:
public int romanToInt(String s) { int nums[]=new int[s.length()]; for(int i=0;i<s.length();i++){ switch (s.charAt(i)){ case 'M': nums[i]=1000; break; case 'D': nums[i]=500; break; case 'C': nums[i]=100; break; case 'L': nums[i]=50; break; case 'X' : nums[i]=10; break; case 'V': nums[i]=5; break; case 'I': nums[i]=1; break; } } /*int sum=0;//从前往后遍历或者下面那种从后往前遍历 for(int i=0;i<nums.length-1;i++){ if(nums[i]<nums[i+1]) sum-=nums[i]; else sum+=nums[i]; } return sum+nums[nums.length-1];*/ int sum=nums[nums.length-1]; for(int i=nums.length-2;i>=0;i--){ if(nums[i]<nums[i+1]) sum-=nums[i]; else sum+=nums[i]; } return sum; }