Skip to content

Latest commit

 

History

History
224 lines (182 loc) · 6.16 KB

File metadata and controls

224 lines (182 loc) · 6.16 KB

English Version

题目描述

罗马数字包含以下七种字符: I, V, X, LCD 和 M

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

例如, 罗马数字 2 写做 II ,即为两个并列的 1。12 写做 XII ,即为 X + II 。 27 写做  XXVII, 即为 XX + V + II 。

通常情况下,罗马数字中小的数字在大的数字的右边。但也存在特例,例如 4 不写做 IIII,而是 IV。数字 1 在数字 5 的左边,所表示的数等于大数 5 减小数 1 得到的数值 4 。同样地,数字 9 表示为 IX。这个特殊的规则只适用于以下六种情况:

  • I 可以放在 V (5) 和 X (10) 的左边,来表示 4 和 9。
  • X 可以放在 L (50) 和 C (100) 的左边,来表示 40 和 90。 
  • C 可以放在 D (500) 和 M (1000) 的左边,来表示 400 和 900。

给你一个整数,将其转为罗马数字。

 

示例 1:

输入: num = 3
输出: "III"

示例 2:

输入: num = 4
输出: "IV"

示例 3:

输入: num = 9
输出: "IX"

示例 4:

输入: num = 58
输出: "LVIII"
解释: L = 50, V = 5, III = 3.

示例 5:

输入: num = 1994
输出: "MCMXCIV"
解释: M = 1000, CM = 900, XC = 90, IV = 4.

 

提示:

  • 1 <= num <= 3999

解法

方法一:贪心

我们可以先将所有可能的符号 $cs$ 和对应的数值 $vs$ 列出来,然后从大到小枚举每个数值 $vs[i]$,每次尽可能多地使用该数值对应的符号 $cs[i]$,直到数字 $num$ 变为 $0$

时间复杂度为 $O(m)$,空间复杂度为 $O(m)$。其中 $m$ 为符号的个数。

class Solution:
    def intToRoman(self, num: int) -> str:
        cs = ('M', 'CM', 'D', 'CD', 'C', 'XC', 'L', 'XL', 'X', 'IX', 'V', 'IV', 'I')
        vs = (1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1)
        ans = []
        for c, v in zip(cs, vs):
            while num >= v:
                num -= v
                ans.append(c)
        return ''.join(ans)
class Solution {
    public String intToRoman(int num) {
        List<String> cs
            = List.of("M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I");
        List<Integer> vs = List.of(1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1);
        StringBuilder ans = new StringBuilder();
        for (int i = 0, n = cs.size(); i < n; ++i) {
            while (num >= vs.get(i)) {
                num -= vs.get(i);
                ans.append(cs.get(i));
            }
        }
        return ans.toString();
    }
}
class Solution {
public:
    string intToRoman(int num) {
        vector<string> cs = {"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"};
        vector<int> vs = {1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1};
        string ans;
        for (int i = 0; i < cs.size(); ++i) {
            while (num >= vs[i]) {
                num -= vs[i];
                ans += cs[i];
            }
        }
        return ans;
    }
};
func intToRoman(num int) string {
	cs := []string{"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"}
	vs := []int{1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1}
	ans := &strings.Builder{}
	for i, v := range vs {
		for num >= v {
			num -= v
			ans.WriteString(cs[i])
		}
	}
	return ans.String()
}
function intToRoman(num: number): string {
    const cs: string[] = ['M', 'CM', 'D', 'CD', 'C', 'XC', 'L', 'XL', 'X', 'IX', 'V', 'IV', 'I'];
    const vs: number[] = [1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1];
    const ans: string[] = [];
    for (let i = 0; i < vs.length; ++i) {
        while (num >= vs[i]) {
            num -= vs[i];
            ans.push(cs[i]);
        }
    }
    return ans.join('');
}
public class Solution {
    public string IntToRoman(int num) {
        List<string> cs = new List<string>{"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"};
        List<int> vs = new List<int>{1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1};
        StringBuilder ans = new StringBuilder();
        for (int i = 0; i < cs.Count; i++) {
            while (num >= vs[i]) {
                ans.Append(cs[i]);
                num -= vs[i];
            }
        }
        return ans.ToString();
    }
}
class Solution {
    /**
     * @param int $num
     * @return string
     */

    function intToRoman($num) {
        $values = [
            'M' => 1000,
            'CM' => 900,
            'D' => 500,
            'CD' => 400,
            'C' => 100,
            'XC' => 90,
            'L' => 50,
            'XL' => 40,
            'X' => 10,
            'IX' => 9,
            'V' => 5,
            'IV' => 4,
            'I' => 1,
        ];

        $result = '';

        foreach ($values as $roman => $value) {
            while ($num >= $value) {
                $result .= $roman;
                $num -= $value;
            }
        }

        return $result;
    }
}