leetcode-java.T012_IntegerToRoman 数字转罗马数字

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每天坚持刷leetcode----- 数字转罗马数字
package leetcode.T012_IntegerToRoman;/**    * @Title: Solution.java  * @Package leetcode.T012_IntegerToRoman  * @Description: TODO  * @author zhouzhixiang   * @date 2017-6-4 上午1:31:40  * @version V1.0    */public class Solution { /**     * <pre>     * Given an integer, convert it to a roman numeral.     *     * Input is guaranteed to be within the range from 1 to 3999.     *     * 罗马数字的表示：     * 个位数举例     * (I, 1)  (II, 2)  (III, 3)  (IV, 4)  (V, 5)  (VI, 6)  (VII, 7)  (VIII, 8)  (IX, 9)     *     * 十位数举例     * (X, 10)  (XI, 11)  (XII, 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)     *     * 百位数举例     * (C, 100)  (CC, 200)  (CCC, 300)  (CD, 400)  (D, 500)  (DC, 600)  (DCC, 700)     * (DCCC, 800)  (CM, 900)  (CMXCIX, 999)     *     * 千位数举例     * (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)     *     * 题目大意：     * 输入一个数字，将它转成一个罗马数字，输入的数字在[1, 3999]之间     *     * </pre>     *     * @param num     * @return     */public static void main(String[] args) {System.out.println(new Solution().intoRoman(199));}// 参考public String intoRoman(int num){String[][] base = new String[][]{                {"I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"}, // 个位的表示                {"X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"}, // 十位的表示                {"C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"}, // 百倍的表示                {"M", "MM", "MMM", "", "", "", "", "", ""}}; // 千位的表示String result = "";// 从个数开始遍历，每遍历一次，result的个位就除去// 同时将被除去的各位转成Romanfor (int i = 0; num!=0; num/=10, i++) {// 如果不为0，说明个位上有值，要进行增加操作if(num%10!=0){// 拼接结果result = base[i][num%10-1]+result;}}return result;}}
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