难度中等878
罗马数字包含以下七种字符: I
, V
, X
, L
,C
,D
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:
代码语言:javascript复制输入: num = 3
输出: "III"
示例 2:
代码语言:javascript复制输入: num = 4
输出: "IV"
示例 3:
代码语言:javascript复制输入: num = 9
输出: "IX"
示例 4:
代码语言:javascript复制输入: num = 58
输出: "LVIII"
解释: L = 50, V = 5, III = 3.
示例 5:
代码语言:javascript复制输入: num = 1994
输出: "MCMXCIV"
解释: M = 1000, CM = 900, XC = 90, IV = 4.
提示:
1 <= num <= 3999
通过次数294,938提交次数445,122
代码语言:javascript复制code
string mm[4][11] ={
{"","I","II","III","IV","V","VI","VII","VIII","IX" },
{"","X","XX","XXX","XL","L","LX","LXX","LXXX","XC"},
{"","C","CC","CCC","CD","D","DC","DCC","DCCC","CM"},
{"","M","MM","MMM"}
};
代码语言:javascript复制class Solution {
public:
string mm[4][11] ={
{"","I","II","III","IV","V","VI","VII","VIII","IX" },
{"","X","XX","XXX","XL","L","LX","LXX","LXXX","XC"},
{"","C","CC","CCC","CD","D","DC","DCC","DCCC","CM"},
{"","M","MM","MMM"}
};
string intToRoman(int num) {
int pos=0;
int idx=0;
//int num=0;
string str[100];
string ans;
while(num!=0){
int cur=num;
if(cur!=0){
str[idx]=mm[pos][cur];
idx ;
}
pos ;
num=num/10;
}
//ans=str[0];
while(idx>=0){
ans =str[idx];
idx--;
}
return ans;
}
};
代码语言:javascript复制#include <stdio.h>
#include <iostream>
using namespace std;
#include <climits>
string intToRoman(int num);
int main(void) {
cout<<intToRoman(588)<<endl;
return 0;
}
string mm[4][11] ={
{"","I","II","III","IV","V","VI","VII","VIII","IX" },
{"","X","XX","XXX","XL","L","LX","LXX","LXXX","XC"},
{"","C","CC","CCC","CD","D","DC","DCC","DCCC","CM"},
{"","M","MM","MMM"}
};
string intToRoman(int num) {
int pos=0;
int idx=0;
//int num=0;
string str[100];
string ans;
while(num!=0){
int cur=num;
if(cur!=0){
str[idx]=mm[pos][cur];
idx ;
}
pos ;
num=num/10;
}
//ans=str[0];
while(idx>=0){
ans =str[idx];
idx--;
}
return ans;
}