Given an integer, convert it to a roman numeral.
Input is guaranteed to be within the range from 1 to 3999.java
罗马数字共有7个,即I(1)、V(5)、X(10)、L(50)、C(100)、D(500)和M(1000)。数组
一、重复数次:一个罗马数字重复几回,就表示这个数的几倍。app
二、右加左减:
2.1 在较大的罗马数字的右边记上较小的罗马数字,表示大数字加小数字。
2.2 在较大的罗马数字的左边记上较小的罗马数字,表示大数字减少数字。
2.3 左减的数字有限制,仅限于I、X、C。好比45不能够写成VL,只能是XLV
2.4 可是,左减时不可跨越一个位数。好比,99不能够用IC(100 - 1)表示,而是用XCIX([100 - 10] + [10 - 1])表示。(等同于阿拉伯数字每位数字分别表示。)
2.5 左减数字必须为一位,好比8写成VIII,而非IIX。
2.6 右加数字不可连续超过三位,好比14写成XIV,而非XIIII。(见下方“数码限制”一项。)ui
三、加线乘千:
3.1 在罗马数字的上方加上一条横线或者加上下标的Ⅿ,表示将这个数乘以1000,便是原数的1000倍。
3.2 同理,若是上方有两条横线,便是原数的1000000(1000^{2})倍。code
四、数码限制:
4.1 同一数码最多只能出现三次,如40不可表示为XXXX,而要表示为XL。
4.2 例外:因为IV是古罗马神话主神朱庇特(即IVPITER,古罗马字母里没有J和U)的首字,所以有时用IIII代替Ⅳ。字符串
千百十个 即便每位有10个数字,一共也就40个数字,用数组枚举便可;it
public String intToRoman(int x) { int tail=x%10; String s=""; //个位 String[] roman1={"","I","II","III","IV","V","VI","VII","VIII","IX"}; s=roman1[tail]+s; //十位 x=x/10; tail=x%10; if(x>0){ String[] roman2={"","X","XX","XXX","XL","L","LX","LXX","LXXX","XC"}; s=roman2[tail]+s; } //百位 x=x/10; tail=x%10; if(x>0){ String[] roman3={"","C","CC","CCC","CD","D","DC","DCC","DCCC","CM"}; s=roman3[tail]+s; } //千位 x=x/10; tail=x%10; if(x>0){ String[] roman4={"","M","MM","MMM"}; s=roman4[tail]+s; } return s; }
改进写法,上面的虽然根据判断,在容许状况下没有建立字符串数组,可是代码看上去很混乱。
如下代码风格更为简洁。class
public 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]; }
public String intToRoman(int num) { int[] values = {1000,900,500,400,100,90,50,40,10,9,5,4,1}; String[] strs = {"M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"}; StringBuilder sb = new StringBuilder(); for(int i=0;i<values.length;i++) { while(num >= values[i]) { num -= values[i]; sb.append(strs[i]); } } return sb.toString(); }