Leetcode question | Integer to Roman

Leetcode question | Integer to Roman

Problem:

 

Given an integer, convert it to a roman numeral.Input is guaranteed to be within the range from 1 to 3999.

Converts an integer from 1 to 3999 to a Roman number.

 

 

Thinking:

(1)

Example

Single digit example
I, 1] II, 2] III, 3] IV, 4] V, 5] vi, 6] VII, 7] VIII, 8] IX, 9]
Ten-digit example
X, 10] average, 11] average, 12] XIII, 13] XIV, 14] XV, 15] XVI, 16] XVII, 17] XVIII, 18] XIX, 19] XX, 20] XXI, 21] XXII, 22] XXIX, 29] XXX, 30] XXXIV, 34] XXXV, 35] XXXIX, 39] XL, 40] L, 50] LI, 51] LV, 55] LX, 60] LXV, 65] LXXX, 80] XC, 90] XCIII, 93] XCV, 95] XCVIII, 98] XCIX, 99]
Example of hundreds of BITs
C, 100] CC, 200] CCC, 300] CD, 400] D, 500] DC, 600] DCC, 700] DCCC, 800] CM, 900] CMXCIX, 999]
Example of a thousand bits
M, 1000] MC, 1100] MCD, 1400] MD, 1500] MDC, 1600] MDCLXVI, 1666] MDCCCLXXXVIII, 1888] MDCCCXCIX, 1899] MCM, 1900] MCMLXXVI, 1976] MCMLXXXIV, 1984] MCMXC, 1990] MM, 2000] MMMCMXCIX, 3999]
 

(2) Tips for avoiding N multi-condition judgment: use the law of the Roman numerals to subtract a base to simplify the complexity.

 

Code:

 

class Solution {public:    string intToRoman(int num)    {        int values[] = {1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1 };        string numerals[] = {"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I" };        string result;        for (int i = 0; i < 13; i++) {            while (num >= values[i]) {                num -= values[i];                result.append(numerals[i]);            }        }        return result;    }};