[leetcode] 13. Roman numeral to integer

13、 Roman numeral to integer

Roman numerals contain the following seven characters : I, V, X, L,C,D and M.

 character            The number 
I             1
V             5
X             10
L             50
C             100
D             500
M             1000

for example , Rome digital 2 Write to do II , Two parallel 1 .12 Write to do XII , That is to say X + II . 27 Write to do XXVII, That is to say XX + V + II .

Usually , The small numbers in roman numbers are to the right of the big ones . But there are special cases , for example 4 Do not write IIII, It is IV. Numbers 1 In number 5 Left side , The number represented is equal to the large number 5 Decimal reduction 1 Value obtained 4 . similarly , Numbers 9 Expressed as IX. This special rule only applies to the following six cases ：

• I Can be placed in V (5) and X (10) Left side , To express 4 and 9.
• X Can be placed in L (50) and C (100) Left side , To express 40 and 90.
• C Can be placed in D (500) and M (1000) Left side , To express 400 and 900.
  Given a Roman number , Convert it to an integer .

Example 1:

 Input : s = "III"
 Output : 3

Example 2:

 Input : s = "IV"
 Output : 4

Example 3:

 Input : s = "IX"
 Output : 9

Example 4:

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

Example 5:

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

Tips ：

1 <= s.length <= 15
s Characters only (‘I’, ‘V’, ‘X’, ‘L’, ‘C’, ‘D’, ‘M’)
Topic data assurance s It's a valid Roman number , And it means that the integer is in the range [1, 3999] Inside
The test cases given in the title all conform to the Roman numeral writing rules , There will be no straddle, etc .
IL and IM Such an example does not meet the requirements of the title ,49 You should write XLIX,999 You should write CMXCIX .
Detailed rules for writing Roman numerals , You can refer to Rome digital - Mathematics .

Their thinking ：

According to the description of the title , The rules can be summarized as follows ：

1. Roman numerals by I,V,X,L,C,D,M constitute ;
2. When the small value is to the left of the large value , Then decrease the value , Such as IV=5-1=4;
3. When the small value is to the right of the large value , Then add a small value , Such as VI=5+1=6;
4. It can be seen from the above that , The right value is always positive , So the last one must be positive .

blunt , Put a small value to the left of the large value , It's subtraction , Otherwise it's addition .

In code implementation , You can look back at one more , Compare the size relationship between the current bit and the next bit , So as to determine whether the current bit is added or subtracted . When there is no next , Just add .

You can also keep the value of the current bit , When traversing to the next bit , Compare the size relationship between the reserved value and the traversal bit , Then determine whether the retention value is plus or minus . The last one can add .

Staff code ：

class Solution {
    
    public int romanToInt(String s) {
    
        int sum = 0;
        int preNum = getValue(s.charAt(0));
        for(int i = 1;i < s.length(); i ++) {
    
            int num = getValue(s.charAt(i));
            if(preNum < num) {
    
                sum -= preNum;
            } else {
    
                sum += preNum;
            }
            preNum = num;
        }
        sum += preNum;
        return sum;
    }
    
    private int getValue(char ch) {
    
        switch(ch) {
    
            case 'I': return 1;
            case 'V': return 5;
            case 'X': return 10;
            case 'L': return 50;
            case 'C': return 100;
            case 'D': return 500;
            case 'M': return 1000;
            default: return 0;
        }
    }
}