# 引言
题目链接:https://leetcode.com/problems/integer-to-roman/description/
# 题目大意
输入一个整型数字,将这个数字转换为罗马数字
有如下约定
Symbol Value
I 1
V 5
X 10
L 50
C 100
D 500
M 1000
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.
- Example
Input: 3
Output: "III"
Input: 4
Output: "IV"
Input: 9
Output: "IX"
Input: 58
Output: "LVIII"
Explanation: C = 100, L = 50, XXX = 30 and III = 3.
Input: 1994
Output: "MCMXCIV"
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.
# 题解
# 一句话题解
一个简单模拟,按照罗马数字的构造方式模拟即可 (**Hint:** 关于 9 或者 4 的特殊情况可以通过手动构造符号映射避免额外的判断)
# 复杂度
时间复杂度 O(n)
空间复杂度 O(n)
# AC 代码
c++
版本
class Solution | |
{ | |
public: | |
string intToRoman(int num) | |
{ | |
vector<pair<int, string>> romanSymbol{ | |
{1000, "M"}, {900, "CM"}, {500, "D"}, {400, "CD"}, | |
{100, "C"}, {90, "XC"}, {50, "L"}, {40, "XL"}, | |
{10, "X"}, {9, "IX"}, {5, "V"}, {4, "IV"}, {1, "I"}}; | |
vector<pair<int, string>>::const_iterator iter = romanSymbol.cbegin(); | |
string ans = ""; | |
while (num > 0 && iter != romanSymbol.cend()) | |
{ | |
if (num >= iter->first) | |
{ | |
num -= iter->first; | |
ans += iter->second; | |
} | |
else | |
{ | |
++iter; | |
} | |
} | |
return ans; | |
} | |
}; |
go
版本
type roman struct { | |
value int | |
symbol string | |
} | |
func intToRoman(num int) string { | |
romans := []roman{ | |
{1000, "M"}, {900, "CM"}, {500, "D"}, {400, "CD"}, | |
{100, "C"}, {90, "XC"}, {50, "L"}, {40, "XL"}, | |
{10, "X"}, {9, "IX"}, {5, "V"}, {4, "IV"}, {1, "I"}} | |
ans := "" | |
index := 0 | |
for num > 0 { | |
if num >= romans[index].value { | |
num -= romans[index].value | |
ans += romans[index].symbol | |
} else { | |
index++ | |
} | |
} | |
return ans | |
} |