赞
踩
本文介绍了 LeetCode 第 13 题 , “Roman to Integer”, 也就是 “罗马数字转整数” 的问题.
本文使用 C# 语言完成题目。
LeetCode 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 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.
LeetCode 13. 罗马数字转整数
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: L = 50, V= 5, III = 3.
示例 5:
输入: “MCMXCIV”
输出: 1994
解释: M = 1000, CM = 900, XC = 90, IV = 4.
方法与第12题类似,只需要从头到尾遍历字符串,并依次转化为数字求和,即可得到最终的值。
参考代码:
public int RomanToInt(string s) { string[] romans = { "M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I" }; int[] nums = { 1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1 }; int result = 0; int sp = 0; //sPointer int rnp = 0; //romans and nums pointer int n = s.Length; while (sp < n) { if (s.IndexOf(romans[rnp], sp) == sp) { result += nums[rnp]; sp += romans[rnp].Length; } else rnp++; } return result; }
执行结果 通过。 执行用时: 100ms, 内存消耗 26.6M
时间复杂度:O(n)
遍历长度为n的字符串。考虑到题目范围为1-3999,也可以认为是O(1).
空间复杂度:O(1)
所用空间大小为固定值。
题目:
Copyright © 2003-2013 www.wpsshop.cn 版权所有,并保留所有权利。