Given an integer, convert it to a roman numeral.
Input is guaranteed to be within the range from 1 to 3999.java
羅馬數字共有7個,即I(1)、V(5)、X(10)、L(50)、C(100)、D(500)和M(1000)。數組
一、重複數次:一個羅馬數字重複幾回,就表示這個數的幾倍。app
二、右加左減:
2.1 在較大的羅馬數字的右邊記上較小的羅馬數字,表示大數字加小數字。
2.2 在較大的羅馬數字的左邊記上較小的羅馬數字,表示大數字減少數字。
2.3 左減的數字有限制,僅限於I、X、C。好比45不能夠寫成VL,只能是XLV
2.4 可是,左減時不可跨越一個位數。好比,99不能夠用IC(100 - 1)表示,而是用XCIX([100 - 10] + [10 - 1])表示。(等同於阿拉伯數字每位數字分別表示。)
2.5 左減數字必須爲一位,好比8寫成VIII,而非IIX。
2.6 右加數字不可連續超過三位,好比14寫成XIV,而非XIIII。(見下方「數碼限制」一項。)ui
三、加線乘千:
3.1 在羅馬數字的上方加上一條橫線或者加上下標的Ⅿ,表示將這個數乘以1000,便是原數的1000倍。
3.2 同理,若是上方有兩條橫線,便是原數的1000000(1000^{2})倍。code
四、數碼限制:
4.1 同一數碼最多隻能出現三次,如40不可表示爲XXXX,而要表示爲XL。
4.2 例外:因爲IV是古羅馬神話主神朱庇特(即IVPITER,古羅馬字母裏沒有J和U)的首字,所以有時用IIII代替Ⅳ。字符串
千百十個 即便每位有10個數字,一共也就40個數字,用數組枚舉便可;it
public String intToRoman(int x) { int tail=x%10; String s=""; //個位 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; }
改進寫法,上面的雖然根據判斷,在容許狀況下沒有建立字符串數組,可是代碼看上去很混亂。
如下代碼風格更爲簡潔。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(); }