012J. Integer to Roman
这道题没有什么意义。
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.
这个解法还是很暴力的。
class Solution {
public static String intToRoman(int num) {
String M[] = {"", "M", "MM", "MMM"};
String C[] = {"", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"};
String X[] = {"", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"};
String I[] = {"", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"};
return M[num/1000] + C[(num%1000)/100] + X[(num%100)/10] + I[num%10];
}
}
这个更好!当然无论如何这个题都不是特别有意思。
class Solution {
private static final int[] values = {1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1};
private static final String[] symbols = {"M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"};
public String intToRoman(int num) {
StringBuilder sb = new StringBuilder();
// Loop through each symbol, stopping if num becomes 0.
for (int i = 0; i < values.length && num > 0; i++) {
// Repeat while the current symbol still fits into num.
while (values[i] <= num) {
num -= values[i];
sb.append(symbols[i]);
}
}
return sb.toString();
}
}
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