12. Integer to Roman
Seven different symbols represent Roman numerals with the following values:
| Symbol | Value |
|---|---|
| I | 1 |
| V | 5 |
| X | 10 |
| L | 50 |
| C | 100 |
| D | 500 |
| M | 1000 |
Roman numerals are formed by appending the conversions of decimal place values from highest to lowest. Converting a decimal place value into a Roman numeral has the following rules:
- If the value does not start with 4 or 9, select the symbol of the maximal value that can be subtracted from the input, append that symbol to the result, subtract its value, and convert the remainder to a Roman numeral.
- If the value starts with 4 or 9 use the subtractive form representing one symbol subtracted from the following symbol, for example, 4 is 1 (
I) less than 5 (V):IVand 9 is 1 (I) less than 10 (X):IX. Only the following subtractive forms are used: 4 (IV), 9 (IX), 40 (XL), 90 (XC), 400 (CD) and 900 (CM). - Only powers of 10 (
I,X,C,M) can be appended consecutively at most 3 times to represent multiples of 10. You cannot append 5 (V), 50 (L), or 500 (D) multiple times. If you need to append a symbol 4 times use the subtractive form.
Given an integer, convert it to a Roman numeral.
Example 1:
Input: num = 3749
Output: "MMMDCCXLIX"
Explanation:
3000 = MMM as 1000 (M) + 1000 (M) + 1000 (M)
700 = DCC as 500 (D) + 100 (C) + 100 (C)
40 = XL as 10 (X) less of 50 (L)
9 = IX as 1 (I) less of 10 (X)
Note: 49 is not 1 (I) less of 50 (L) because the conversion is based on decimal places
Example 2:
Input: num = 58
Output: "LVIII"
Explanation:
Example 3:
Input: num = 1994
Output: "MCMXCIV"
Explanation:
Constraints:
1 <= num <= 3999
Solution:
把 num 拆分成千位数、百位数、十位数和个位数，分别用罗马数字表示。
例如示例 1 的 num=3749，拆分后对应的罗马数字分别为 MMM,DCC,XL,IX。
根据题意，从千位数到个位数，阿拉伯数字与罗马数字的转换关系为：
千位从 1 到 3 依次为 M,MM,MMM。 百位从 1 到 9 依次为 C,CC,CCC,CD,D,DC,DCC,DCCC,CM。 十位从 1 到 9 依次为 X,XX,XXX,XL,L,LX,LXX,LXXX,XC。 个位从 1 到 9 依次为 I,II,III,IV,V,VI,VII,VIII,IX。 把 num 拆分成各个数位的公式为：
千位：\(⌊\frac{1000}{num}⌋\)。例如 \(\lfloor \frac{1000}{3749} \rfloor\)
百位：\(\lfloor \frac{100}{num} \rfloor\) 例如 \(\lfloor \frac{3749}{100} \rfloor \mod 10 = 37 \mod 10 = 7\)
十位：\(\lfloor \frac{num}{10} \rfloor \mod 10\) 例如 $\lfloor \frac{3749}{10} \rfloor \mod 10 = 4 $
个位：\(num \mod 10\)。例如 \(3749 \mod 10 = 9\)。 用字符串数组存储罗马数字，用上述公式计算出的数位当作数组下标，取出对应的罗马数字（字符串）。把这些字符串拼接起来，得到答案。
class Solution {
private static final String[][] R = new String[][]{
{"", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"},
{"", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"},
{"", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"},
{"", "M", "MM", "MMM"}
};
public String intToRoman(int num) {
return R[3][num/1000] + R[2][num / 100 % 10] + R[1][num / 10 % 10] + R[0][num % 10];
}
}
// TC: O(n)
// SC: O(1)
I = 1
II = 2
III = 3
IV = 4 (-1 + 5)
V = 5
VI = 6
VII = 7
VIII = 8
IX = 9 (- 1 + 10)
X = 10
...
L = 50
C = 100
D = 500
M = 1000
class Solution {
public String intToRoman(int num) {
Map<Integer, String> map = new HashMap<>();
map.put(1, "I");
map.put(4, "IV");
map.put(5, "V");
map.put(9, "IX");
map.put(10, "X");
map.put(40, "XL"); // (-10 + 50)
map.put(50, "L");
map.put(90, "XC");
map.put(100, "C");
map.put(400, "CD");
map.put(500, "D");
map.put(900, "CM");
map.put(1000, "M");
int[] values = {1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1};
StringBuilder sb = new StringBuilder();
for (int v : values){
while(num >= v){
sb.append(map.get(v));
num = num - v;
}
}
return sb.toString();
}
}
/*
Symbol Value
I 1
V 5
X 10
L 50
C 100
D 500
M 1000
num = 3749
M M. M D C C. X L I. X
1000 + 1000 + 1000 + 500 + 100 + 100 + 10 + 50 + 1 + 10
s
f
if ( f > s) - > cur. + (f -
*/
// TC: O(n)
// SC: O(1)
class Solution {
public String intToRoman(int num) {
Map<Integer, String> map = new HashMap<>();
map.put(1, "I");
map.put(4, "IV");
map.put(5, "V");
map.put(9, "IX");
map.put(10, "X");
map.put(40, "XL"); // (-10 + 50)
map.put(50, "L");
map.put(90, "XC");
map.put(100, "C");
map.put(400, "CD");
map.put(500, "D");
map.put(900, "CM");
map.put(1000, "M");
StringBuilder sb = new StringBuilder();
int time = 1000;
// 3749
while(time > 0){
int curD = num;
num = num;
if (time != 0){
curD = num / time; // 3749 /1000 = 3
num = num % time; // 749
}
if (curD == 0){
time = time / 10;
continue;
}
/*
If the value does not start with 4 or 9, select the symbol of the maximal value that can be subtracted from the input, append that symbol to the result, subtract its value, and convert the remainder to a Roman numeral.
*/
if (curD != 4 && curD != 9){
if (curD == 5){
sb.append(map.get(5 * time));
}
if (curD < 4){
for (int i = 0; i < curD; i++){
sb.append(map.get(time));// 3
// 1000
}
}
if (curD > 5){
sb.append(map.get(5 * time));
for (int i = 0; i < curD - 5; i++){
sb.append(map.get(time));
}
}
time = time / 10;
}
if (curD == 4){
sb.append(map.get(4 * time));
time = time / 10;
continue;
}
if (curD == 9){
sb.append(map.get(9 * time));
time = time / 10;
continue;
}
}
return sb.toString();
}
}
/*
Symbol Value
I 1
V 5
X 10
L 50
C 100
D 500
M 1000
num = 3749
M M. M D C C. X L I. X
1000 + 1000 + 1000 + 500 + 100 + 100 + 10 + 50 + 1 + 10
s
f
if ( f > s) - > cur. + (f -
*/
class Solution {
public String intToRoman(int num) {
Map<Integer, String> map = new HashMap<>();
map.put(1, "I");
map.put(4, "IV");
map.put(5, "V");
map.put(9, "IX");
map.put(10, "X");
map.put(40, "XL"); // (-10 + 50)
map.put(50, "L");
map.put(90, "XC");
map.put(100, "C");
map.put(400, "CD");
map.put(500, "D");
map.put(900, "CM");
map.put(1000, "M");
StringBuilder sb = new StringBuilder();
int time = 1000;
// 3749
while(time > 0){
int curD = num / time;
num = num % time;
if (curD == 0){
time = time / 10;
continue;
}
if (curD == 4){
sb.append(map.get(4 * time));
time = time / 10;
continue;
}else if (curD == 9){
sb.append(map.get(9 * time));
time = time / 10;
continue;
}else{
if (curD >= 5){
sb.append(map.get(5 * time));
curD = curD - 5;
}
for (int i = 0; i < curD; i++){
sb.append(map.get(time));
}
time = time / 10;
}
}
return sb.toString();
}
}
// TC: O(n)
// SC: O(1)