C# 写 LeetCode easy #13 Roman to Integer

时间:2023-12-25 21:20:37

13、Roman to Integer

Roman numerals are represented by seven different symbols: IVXLCD 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 a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999.

Example 1:

Input: "III"
Output: 3

Example 2:

Input: "IV"
Output: 4

Example 3:

Input: "IX"
Output: 9

Example 4:

Input: "LVIII"
Output: 58
Explanation: L = 50, V= 5, III = 3.

Example 5:

Input: "MCMXCIV"
Output: 1994
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4. 代码:
static void Main(string[] args)
{
string str = "LVIV";
int res=RomanToInteger(str);
Console.WriteLine(res);
Console.ReadKey();
} private static int RomanToInteger(string str)
{
int res = ;
Dictionary<char, int> dic=new Dictionary<char, int> { { 'I', }, { 'V', }, { 'X', }, { 'L', }, { 'C', }, { 'D', }, { 'M', } };
for (int i = ; i < str.Length; ++i)
{
int val = dic[str[i]];
if (i == str.Length - || dic[str[i + ]] <= dic[str[i]])
{
res += val;
}
else
{
res -= val;
}
}
return res;
}

解析:

输入:字符串

输出:整数

思想

  首先,分别将单个罗马数和其所对应的整数存入字典中。

  其次,对于输入的罗马数,将其看作字符串。设置目前数为0,开始遍历,根据规律,从第一个字符到倒数第二个字符,每个字符在字典中的值与后一个字符比较,若前者大于后者,说明是类似于IV一样的,需要用目前的数减去这个值。否则,用目前的数加上这个值。若循环到最后一个字符,则其在字典中的值直接相加,直到循环结束。

  最后,返回结果。

时间复杂度:O(n)