Leetcode---Integer to Roman

Title:Given An integer, convert it to a Roman numeral.
Input is guaranteed to being within the range from 1 to 3999.
Main topic:Enter an integer number, convert it to Roman numerals, and range between 1--3999. Ideas:before doing this question deliberately Google a little Roman figure is what ghosts, because previously only know the Roman numeral 1---10 (=, =), the wikipedia explanation paste as follows:

　　There are 7 Roman numerals, namely I (1), V (5), X (10), L (50), C (100), D (500) and M (1000). Any positive integer can be represented by the following rules. It is important to note that there is no "0" in the Roman numerals, regardless of the rounding system . Roman numerals are generally considered to be used only for counting , not for calculation.

• Repeat several times: a Roman number repeats several times, indicating several times the number.

• Right plus left minus:
  • The smaller Roman numerals on the right side of the larger Roman numerals indicate large numbers plus small numbers.
  • However, the left subtraction cannot span a single bit value. For example, 99 cannot be represented by an IC (), but by XCIX (). (equivalent to each digit of the Arabic numerals.) )

• add line by thousand:
  • Similarly, if there are two horizontal lines above, That is 1000000 () times the original number.

• Digital Restrictions:
  • The same digital can only appear three consecutive times, such as 40 cannot be represented as XXXX, but to be represented as XL.
  • Exception: Since IV is the first word of the Roman mythology Lord Jupiter (ie, ivpiter, the Roman alphabet does not have J and u), it is sometimes substituted with IIII for IV.

　　　 know what is the Roman number is good to do, my idea is: anyway the largest number until 3999, simply build a table, will be 1--9, ten---90,---900,---3000 in accordance with the subscript corresponding to deposit into a two-dimensional array inside, and then each bit of the number of separate, look up the table , append to the result string. Code:

Class Solution {public:    std::string inttoroman (int num) {        std::string result = "";        std::string romannums[4][10] = {            {"", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"},            {"", "X", "xx", "xx  X "," XL "," L "," LX "," LXX "," LXXX "," XC "},            {" "," C "," CC "," CCC "," CD "," D "," DC "," DCC "," DCCC "," CM "},            {" "," M ", "MM", "MMM"}        ;                Remove the first three digits for        (int i = n, j = 3; I! = 1; I/=,--j) {            result + = romannums[j][num/i%];        }        Take out the last digit of the number        result + = romannums[0][num%];                return result;}    ;

The fastest execution time for the above code in Leetcode is 36ms, but changing the two-dimensional array to char * is the fastest execution time is 24ms ... Description of C + + string here is no char* fast ah.


Reference: https://zh.wikipedia.org/wiki/%E7%BD%97%E9%A9%AC%E6%95%B0%E5%AD%97

Leetcode---Integer to Roman