LeetCode 12. Integer to Roman

原题链接在这里：https://leetcode.com/problems/integer-to-roman/

题目：

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

Example 1:

Input: 3
Output: "III"

Example 2:

Input: 4
Output: "IV"

Example 3:

Input: 9
Output: "IX"

Example 4:

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

Example 5:

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

题解：

每次从最大的数开始. 之所以需要考虑900, 400, 90, 40这些情况, 而不考虑800, 300, 200, 80, 30, 20这些情况是因为, 只有左侧加有特罗马字符的才需要特殊添加.

右侧加罗马字符可以等下下一次正常添加。

e.g. num = 8, 加"V"，然后剩下3，依次加三个"I"就可以了.

Time Complexity: O(sb.length() * values.length), 因为扫了一遍digit数组，扫了一遍roman数组.

Space: O(sb.length()).

AC Java:

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

类似Roman to Integer.