【LeetCode】012.Integer to Roman

来源：互联网发布：网络链路冗余技术编辑：程序博客网时间：2024/06/05 11:17

题目：

Given an integer, convert it to a roman numeral.

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

解答：

从千位到个位，根据位数不同选择不同的字母，char1,char5,char10分别代表该位表示1，5和10的字母。

c为当前位数字（0---9），模式为：

c -- 9 char1 + char10

c :5 ---8 char5 + char1[c-5]

c:4 char1+char5

c:1--3 char1[c]

代码：

public class Solution {    public String intToRoman(int num) {StringBuilder str = new StringBuilder();char[] chars;char char1 = 0, char5=0,char10=0;int mode = 1000;while (mode > 0) {int c = num / mode;num %= mode;switch (mode) {case 1000:char1 = 'M';if(c > 3)return "error";break;case 100:char1 = 'C';char5 = 'D';char10 = 'M';break;case 10:char1 = 'X';char5 = 'L';char10 = 'C';break;case 1:char1 = 'I';char5 = 'V';char10 = 'X';break;default:break;}if (c == 9)str.append(char1).append(char10);else if (c >= 5) { // 5 --- 8chars = new char[c - 5 + 1];Arrays.fill(chars, char1);chars[0] = char5;str.append(chars);} else if (c == 4) // 4str.append(char1).append(char5);else if(c > 0){ // 1 --- 3chars = new char[c];Arrays.fill(chars, char1);str.append(chars);}mode /= 10;}return str.toString();}}

0 0