[LeetCode] 13. Roman to Integer 罗马数字转为整数

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.

12. Integer to Roman 的拓展，反过来转换，将罗马数字转为整数。

解法：对比s[i]和s[i+1]的罗马数字对应的数字的大小，如果s[i]小于s[i+1]则是出现9的情况，减去此数值。其它情况均累加相对应的数值。

Java:

public int romanToInt(String s) {
     int sum=0;
    if(s.indexOf("IV")!=-1){sum-=2;}
    if(s.indexOf("IX")!=-1){sum-=2;}
    if(s.indexOf("XL")!=-1){sum-=20;}
    if(s.indexOf("XC")!=-1){sum-=20;}
    if(s.indexOf("CD")!=-1){sum-=200;}
    if(s.indexOf("CM")!=-1){sum-=200;}
    
    char c[]=s.toCharArray();
    int count=0;
    
   for(;count<=s.length()-1;count++){
       if(c[count]=='M') sum+=1000;
       if(c[count]=='D') sum+=500;
       if(c[count]=='C') sum+=100;
       if(c[count]=='L') sum+=50;
       if(c[count]=='X') sum+=10;
       if(c[count]=='V') sum+=5;
       if(c[count]=='I') sum+=1;
       
   }
   
   return sum;
    
}

Java:

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;
    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];
}　　

Python:

class Solution:
# @param {string} s
# @return {integer}
def romanToInt(self, s):
    roman = {'M': 1000,'D': 500 ,'C': 100,'L': 50,'X': 10,'V': 5,'I': 1}
    z = 0
    for i in range(0, len(s) - 1):
        if roman[s[i]] < roman[s[i+1]]:
            z -= roman[s[i]]
        else:
            z += roman[s[i]]
    return z + roman[s[-1]]　

Python:

class Solution:
    # @return an integer
    def romanToInt(self, s):
        numeral_map = {"I": 1, "V": 5, "X": 10, "L": 50, "C":100, "D": 500, "M": 1000}
        decimal = 0
        for i in xrange(len(s)):
            if i > 0 and numeral_map[s[i]] > numeral_map[s[i - 1]]:
                decimal += numeral_map[s[i]] - 2 * numeral_map[s[i - 1]]
            else:
                decimal += numeral_map[s[i]]
        return decimal

Python:

def romanToInt(self, s):
        """
        :type s: str
        :rtype: int
        """
        _dict = {'I':1,'V':5,'X':10,'L':50,'C':100,'D':500,'M':1000}
        prev = 0
        sum = 0
        for i in s[::-1]:
            curr = _dict[i]
            if prev > curr:
                sum -= curr
            else:
                sum += curr
            prev = curr
        return sum　　

C++:

class Solution {
public:
    int romanToInt(string s) {
        unordered_map<char, int> numeral_map = {{'I',    1}, {'V',   5}, {'X',  10},
                                                {'L',   50}, {'C', 100}, {'D', 500},
                                                {'M', 1000}};
        int decimal = 0;
        for (int i = 0; i < s.length(); ++i) {
            if (i > 0 && numeral_map[s[i]] > numeral_map[s[i - 1]]) {
                decimal += numeral_map[s[i]] - 2 * numeral_map[s[i - 1]];
            } else {
                decimal += numeral_map[s[i]];
            }
        }
        return decimal;
    }
};

　

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[LeetCode] 12. Integer to Roman 整数转为罗马数字

 

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