[Swift]LeetCode13.罗马数字转整数|RomantoInteger

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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 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: C = 100, L = 50, XXX = 30 and III = 3.

Example 5:

Input: "MCMXCIV"
Output: 1994
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

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。

给定一个罗马数字，将其转换成整数。输入确保在 1 到 3999 的范围内。

示例 1:

输入: "III"
输出: 3

示例 2:

输入: "IV"
输出: 4

示例 3:

输入: "IX"
输出: 9

示例 4:

输入: "LVIII"
输出: 58
解释: C = 100, L = 50, XXX = 30, III = 3.

示例 5:

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

 1 class Solution {
 2     func romanToInt(_ s: String) -> Int {
 3         //用字典存对照表,单引号字符串查找使用双引号
 4         var map:[Character:Int]=["I":1,"V":5,"X":10,"L":50,"C":100,"D":500,"M":1000]
 5         //定义局部变量
 6         var ret :Int = 0
 7         //判断字符串是否为空
 8         if !s.isEmpty
 9         {
10             //遍历字符串,indices属性可以访问字符串中各个字符的所有索引。
11             for keys:String.Index in s.indices
12             {
13                 //注意判断keys的范围只能到倒数第二位
14                 if   keys < s.index(before: s.endIndex) 
15                      && map[s[keys]]! < map[s[s.index(after: keys)]]!
16                 {
17                     ret -= map[s[keys]]!
18                 }
19                 else
20                 {
21                     ret += map[s[keys]]!
22                 }
23             }
24         }
25         return ret
26     }
27 }

高效率版

 1 class Solution {
 2     func romanToInt(_ s: String) -> Int {
 3         var count = 0
 4         var index = s.startIndex
 5         while index != s.endIndex {
 6             let c = s[index]
 7             if c == "I" {
 8                 if s.index(after: index) != s.endIndex && s[s.index(after: index)] == "V" {
 9                     count += 4
10                     index = s.index(after: index)
11                 } else if s.index(after: index) != s.endIndex && s[s.index(after: index)] == "X" {
12                     count += 9
13                     index = s.index(after: index)
14                 } else {
15                     count += 1
16                 }
17                 index = s.index(after: index)
18             } else if c == "X" {
19                 if s.index(after: index) != s.endIndex && s[s.index(after: index)] == "L" {
20                     count += 40
21                     index = s.index(after: index)
22                 } else if s.index(after: index) != s.endIndex && s[s.index(after: index)] == "C" {
23                     count += 90
24                     index = s.index(after: index)
25                 } else {
26                     count += 10
27                 }
28                 index = s.index(after: index)
29             } else if c == "C" {
30                 if s.index(after: index) != s.endIndex && s[s.index(after: index)] == "D" {
31                     count += 400
32                     index = s.index(after: index)
33                 } else if s.index(after: index) != s.endIndex && s[s.index(after: index)] == "M" {
34                     count += 900
35                     index = s.index(after: index)
36                 } else {
37                     count += 100
38                 }
39                 index = s.index(after: index)
40             } else if c == "V" {
41                 count += 5
42                 index = s.index(after: index)
43             } else if c == "L" {
44                 count += 50
45                 index = s.index(after: index)
46             } else if c == "D" {
47                 count += 500
48                 index = s.index(after: index)
49             } else if c == "M" {
50                 count += 1000
51                 index = s.index(after: index)
52             }
53         }
54         return count
55     }
56 }