LeetCode 12. Integer to Roman 整数转罗马数字（Java）

题目：

Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.

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.

解答：

起初没有想到简单的解决方法，因此采用了大量的 if-else 进行分情况讨论。从高位开始逐位进行转换。

class Solution {
    public String intToRoman(int num) {
        StringBuffer str = new StringBuffer();
        int temp = 0;
        if( num < 4000 && num > 999 ) {
            temp = num/1000;
            for(int i=0; i<temp; i++) {
                str.append("M");
            }
            num %= 1000; 
        }
        if( num < 1000 && num > 99 ) {
            temp = num/100;
            if( temp < 4 ){
                for(int i=0; i<temp; i++){
                    str.append("C");
                }
            } else if( temp == 4) {
                str.append("CD");
            } else if( temp > 4 && temp < 9) {
                str.append("D");
                temp -= 5;
                for(int i=0; i<temp; i++) {
                    str.append("C");
                }
            } else if( temp==9 ) {
                str.append("CM");
            }
            num %= 100; 
        }
        if( num < 100 && num > 9 ) {
            temp = num/10;
            if( temp < 4 ){
                for(int i=0; i<temp; i++){
                    str.append("X");
                }
            } else if( temp == 4) {
                str.append("XL");
            } else if( temp > 4 && temp < 9) {
                str.append("L");
                temp -= 5;
                for(int i=0; i<temp; i++) {
                    str.append("X");
                }
            } else if( temp == 9) {
                str.append("XC");
            }
            num %= 10; 
        }
        if( num < 10 && num >0 ) {
            temp = num;
            if( temp < 4) {
                for( int i=1; i<=temp; i++ ) {
                    str.append("I");
                }
            } else if ( temp == 4) {
                str.append("IV");
            }else if ( temp < 9 && temp > 4) {
                str.append("V");
                temp -= 5;
                for( int i=1; i<=temp; i++ ) {
                    str.append("I");
                }
            }else if( temp == 9 ) {
                str.append("IX");
            }
        }
        return str.toString();
    }
}

后来看了他人的解法，对上面解法进行的优化，使得代码变得及其整洁。

class Solution {
    public String intToRoman(int num) {
        String str = "";
        String[] roman = {"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"};
        int[] value = {1000, 900, 500, 400,100, 90, 50, 40, 10, 9, 5, 4, 1};
        for(int i=0; i<value.length; i++) {
            while(num>=value[i]) {
                str += roman[i];
                num -= value[i];
            }
        }
        return str;
    }
}