1.問(wèn)題描述

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.

來(lái)自 https://leetcode.com/problems/roman-to-integer/description/

2.題目分析

題目要求把羅馬數(shù)字轉(zhuǎn)換成阿拉伯?dāng)?shù)字，首先我們先來(lái)看看什么是羅馬數(shù)字，題目中已經(jīng)給出了具體的計(jì)數(shù)方法的描述。我們先分析這個(gè)羅馬數(shù)字的字符串，這個(gè)字符串可以劃分為千位，百位，十位，個(gè)位，讀出千位后，后面緊接著就是百位了，以此類推。我的想法就是建一個(gè)二維的表，分別存儲(chǔ)個(gè)位，十位，百位，千位。先從千位找起，找到后再找百位，然后十位，個(gè)位，找到的條件是查找的起點(diǎn)和find函數(shù)的的返回一致(比如4-IV 和5-v,當(dāng)我們拿v去匹配時(shí)會(huì)找到iv才結(jié)束，因此不能用find()!=-1)。每找到一個(gè)，我們就要更新下一次查找的字符串的起點(diǎn)。

3.C++代碼

//我的代碼：（beats 34%） int romanToInt(string s) {//對(duì)照表char *c[4][10] = {{ "","I","II","III","IV","V","VI","VII","VIII","IX" },{ "","X","XX","XXX","XL","L","LX","LXX","LXXX","XC" },{ "","C","CC","CCC","CD","D","DC","DCC","DCCC","CM" },{ "","M","MM","MMM" }};int res = 0;//int i = 3;int j = 3;int flag = 0;for ( i; i >= 0; i--){for ( j ; j > 0; j--){if (s.find(c[i][j],flag) == flag)//在起點(diǎn)處找到匹配的{res += j*pow(10, i);string tmp_str = c[i][j];flag+=tmp_str.length();//更新下一次查找的起點(diǎn)break;}}j = 9;} return res; } //討論區(qū)比較好的方法 int romanToInt2(string s) {int res = 0;for (int i = s.length() - 1; i >= 0; i--){switch (s[i]){case 'I':res += (res > 5 ? -1 : 1);break;case 'V':res += 5;break;case'X':res += 10 * (res > 50 ? -1 : 1);break;case 'L':res += 50;break;case 'C':res += 100 * (res > 500 ? -1 : 1);break;case 'D':res += 500;break;case 'M':res += 1000;break;}}return res; } //附加阿拉伯轉(zhuǎn)羅馬數(shù)字 string intTointroman(int nums) {string s;char *c[4][10] = { { "","I","II","III","IV","V","VI","VII","VIII","IX" },{ "","X","XX","XXX","XL","L","LX","LXX","LXXX","XC" },{ "","C","CC","CCC","CD","D","DC","DCC","DCCC","CM" },{ "","M","MM","MMM" }};s.append(c[3][nums / 1000]);s.append(c[2][nums % 1000 / 100]);s.append(c[1][nums % 100 / 10]);s.append(c[0][nums % 10]);return s; }

4.相關(guān)知識(shí)點(diǎn)

string類:
https://blog.csdn.net/hero_myself/article/details/52313617

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