LeetCode-12：Integer to Roman(整数转罗马数字)

题目：

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.

例子:

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: C = 100, L = 50, XXX = 30 and III = 3.

Example 5:

Input: 1994
Output: "MCMXCIV"
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

问题解析：

给定一个整数，将其转为罗马数字。输入确保在 1 到 3999 的范围内.

思路标签

分段讨论、贪婪算法

解答：

1.分段讨论

• 建立罗马符号和对应整数的数组；
• 分别讨论当前位（从高位到低位）的大小的不同情况：以x为例。
• 当x<4的情况；当x == 4的情况；当x>4 & x<9的情况；以及当x == 9的情况。

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 += roman[n];
            } 
            else if (x == 4) 
                res = res + roman[n] + roman[n - 1];
            else if (x > 4 && x < 9) {
                res += roman[n - 1];
                for (int i = 6; i <= x; ++i) res += roman[n];
            }
            else if (x == 9) 
                res = res + roman[n] + roman[n - 2];

            num %= value[n];            
        }
        return res;
    }
};

2.贪婪算法

• 建立罗马符号和对应整数的数组；
• 每次查表找出当前最大的数，减去该数后再继续查表。

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