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➤微信公众号:山青咏芝(shanqingyongzhi)➤博客园地址:山青咏芝()➤GitHub地址:➤原文地址: ➤如果链接不是山青咏芝的博客园地址,则可能是爬取作者的文章。➤原文已修改更新!强烈建议点击原文地址阅读!支持作者!支持原创!★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★Roman numerals are represented by seven different symbols: I
, V
, X
, L
, C
, D
and M
.
Symbol ValueI 1V 5X 10L 50C 100D 500M 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: 58Explanation: C = 100, L = 50, XXX = 30 and III = 3.
Example 5:
Input: "MCMXCIV"Output: 1994Explanation: 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 1V 5X 10L 50C 100D 500M 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.indices12 {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 else20 {21 ret += map[s[keys]]!22 }23 }24 }25 return ret26 }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 += 410 index = s.index(after: index)11 } else if s.index(after: index) != s.endIndex && s[s.index(after: index)] == "X" {12 count += 913 index = s.index(after: index)14 } else {15 count += 116 }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 += 4021 index = s.index(after: index)22 } else if s.index(after: index) != s.endIndex && s[s.index(after: index)] == "C" {23 count += 9024 index = s.index(after: index)25 } else {26 count += 1027 }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 += 40032 index = s.index(after: index)33 } else if s.index(after: index) != s.endIndex && s[s.index(after: index)] == "M" {34 count += 90035 index = s.index(after: index)36 } else {37 count += 10038 }39 index = s.index(after: index)40 } else if c == "V" {41 count += 542 index = s.index(after: index)43 } else if c == "L" {44 count += 5045 index = s.index(after: index)46 } else if c == "D" {47 count += 50048 index = s.index(after: index)49 } else if c == "M" {50 count += 100051 index = s.index(after: index)52 }53 }54 return count55 }56 }