C# 写 LeetCode easy #13 Roman to Integer
13、Roman to Integer
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 beforeV
(5) andX
(10) to make 4 and 9.X
can be placed beforeL
(50) andC
(100) to make 40 and 90.C
can be placed beforeD
(500) andM
(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 = 0; Dictionary<char, int> dic=new Dictionary<char, int> { { 'I', 1}, { 'V', 5}, { 'X', 10}, { 'L', 50}, { 'C', 100}, { 'D', 500}, { 'M', 1000} }; for (int i = 0; i < str.Length; ++i) { int val = dic[str[i]]; if (i == str.Length - 1 || dic[str[i + 1]] <= 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)