[LeetCode]12. Integer to Roman整数转罗马数字

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

Symbol       Value
I             1
V             5
X             10
L             50
C             100
D             500
M             1000

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.

题目要转换阿拉伯数字和罗马数字，观察到题目内输入范围较小，在4位数以内，可以直接给出各位映射

class Solution {
    public String intToRoman(int num) {
        String[] v1= {"", "M", "MM", "MMM"};
        String[] v2= {"", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"};
        String[] v3= {"", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"};
        String[] v4= {"", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"};
        return v1[num / 1000] + v2[(num % 1000) / 100] + v3[(num % 100) / 10] + v4[num % 10];
    }
}

上面的方法有点取巧，我们来看看正常的做法

字符          数值
I             1
V             5
X             10
L             50
C             100
D             500
M             1000

假设字符串S=abcd,我们要求出每一位的表示方式，依次要除以1000,100,10，中间要夹杂着500,50,5之类的，所以我们循环的时候每次要+2除下来的结果我们无非是{0,1,2,3,4,5,6,7,8,9}，根据题目意思，4和9是比较特殊的情况，我们可以把结果集分为x<4,x==4,x>4&&x<9,x==9x<4的情况下，x>1才会有转换，并且x是多少就重复几次，可以用for(int i=1;i<4;i++)这里i必须要从1开始x==4的情况下，分两部分加起来，一个5一个1就行了x>4&&x<9的情况下，x>=5需要先把5的字符加上去，再从6开始循环，x>5就再往上加1,for(int i=6;i<=x;i++)x==9的情况下，也是特殊情况，前一位和1加在一起代码如下

class Solution {
    public String intToRoman(int num) {
        String res = "";
        char roman[] = {‘M‘, ‘D‘, ‘C‘, ‘L‘, ‘X‘, ‘V‘, ‘I‘};
        int value[] = {1000, 500, 100, 50, 10, 5, 1};

        for (int n = 0; n < 7; n += 2) {
            int x= num/value[n];
            if (x < 4) {
                for (int i = 1; i <= x; ++i) {
                    res=res+roman[n];
                }
            } else if (x == 4) {
                res=res+roman[n]+roman[n - 1];
            }
            else if (x > 4 && x < 9) {
                res=res+roman[n - 1];
                for (int i = 6; i <= x; ++i) {
                    res=res+roman[n];
                }
            }
            else if (x == 9) {
                res =res+roman[n]+roman[n - 2];
            }
            num=num%value[n];
        }
        return res;
    }
}

原文地址：https://www.cnblogs.com/jchen104/p/10205576.html