12. Integer to Roman

题目：

Given an integer, convert it to a roman numeral.

Input is guaranteed to be within the range from 1 to 3999.

链接：https://leetcode.com/problems/integer-to-roman/#/description

4/9/2017

18%, 115ms

 1 public class Solution {
 2     public String intToRoman(int num) {
 3         HashMap<Integer, Character> hm = new HashMap<Integer, Character>();
 4         StringBuilder sb = new StringBuilder();
 5         hm.put(1, 'I');
 6         hm.put(5, 'V');
 7         hm.put(10, 'X');
 8         hm.put(50, 'L');
 9         hm.put(100, 'C');
10         hm.put(500, 'D');
11         hm.put(1000, 'M');
12 
13         while (num > 0) {
14             if (num / 1000 != 0) {
15                 int count = num / 1000;
16                 num -= 1000 * count;
17                 while (count > 0) {
18                     sb.append(hm.get(1000));
19                     count--;
20                 }
21             } else if (num / 900 != 0) {
22                 sb.append(hm.get(100));
23                 sb.append(hm.get(1000));
24                 num -= 900;
25             } else if (num / 500 != 0) {
26                 sb.append(hm.get(500));
27                 num -= 500;
28             } else if (num / 400 != 0) {
29                 sb.append(hm.get(100));
30                 sb.append(hm.get(500));
31                 num -= 400;
32             } else if (num / 100 != 0) {
33                 int count = num / 100;
34                 num -= 100 * count;
35                 while (count > 0) {
36                     sb.append(hm.get(100));
37                     count--;
38                 }
39             } else if (num / 90 != 0) {
40                 sb.append(hm.get(10));
41                 sb.append(hm.get(100));
42                 num -= 90;
43             } else if (num / 50 != 0) {
44                 sb.append(hm.get(50));
45                 num -= 50;
46             } else if (num / 40 != 0) {
47                 sb.append(hm.get(10));
48                 sb.append(hm.get(50));
49                 num -= 40;
50             } else if (num / 10 != 0) {
51                 int count = num / 10;
52                 num -= 10 * count;
53                 while (count > 0) {
54                     sb.append(hm.get(10));
55                     count--;
56                 }
57             } else if (num / 9 != 0) {
58                 sb.append(hm.get(1));
59                 sb.append(hm.get(10));
60                 num -= 9;
61             } else if (num / 5 != 0) {
62                 sb.append(hm.get(5));
63                 num -= 5;
64             } else if (num / 4 != 0) {
65                 sb.append(hm.get(1));
66                 sb.append(hm.get(5));
67                 num -= 4;
68             } else {
69                 while (num > 0) {
70                     sb.append(hm.get(1));
71                     num--;
72                 }
73             }
74         }
75         return sb.toString();
76     }
77 }

其实可以多列出来很多case，比如9，4的情况。甚至可以列出来所有的例子

别人的答案：

 1 public class Solution {
 2     public String intToRoman(int num) {   //Roman:   I = 1,  V = 5,   X = 10,   L = 50,   C = 100,  D = 500,  M = 1000 
 3         StringBuilder result = new StringBuilder();
 4         String [] symbol = {"M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"};    
 5         int [] value = {1000,900,500,400, 100, 90,  50, 40,  10, 9,   5,  4,   1};   
 6         for(int i = 0; num != 0; i++){
 7             while(num >= value[i]){
 8                 num -= value[i];
 9                 result.append(symbol[i]);
10             }
11         }
12         return result.toString();
13     }
14 }

https://discuss.leetcode.com/topic/12384/simple-solution

1 public static String intToRoman(int num) {
2     String M[] = {"", "M", "MM", "MMM"};
3     String C[] = {"", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"};
4     String X[] = {"", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"};
5     String I[] = {"", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"};
6     return M[num/1000] + C[(num%1000)/100] + X[(num%100)/10] + I[num%10];
7 }