本文共 4605 字，大约阅读时间需要 15 分钟。
题目描述
Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.
罗马数字是用七个不同的标志表示：I, V, X, L, C, D 和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.
例如:2在罗马数字中使用II表示，仅仅是两个1相加。12写作XII，仅仅是X和II相加。27写作XXVII，就是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:
罗马数字常常是由左到右，按照最大到最小的方式去写的。然而，数字4并不是IIII。而是，数字4写为IV。因为1在左边表示减去1，5减1，就是4.同样的原则也适用于9，写作IX。下面有六个使用减法的例子。
* 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.
给你一个罗马数字，将之转化成整型数字。输入的数字是在1至3999范围内的
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.
思路分析
- 按照由简入繁的思路进行思考和分析
- 假设没有减法，从尾部项头部开始遍历所有的字符串的字母，遇到对应的字母就在结果上加上对应的数字
- 有减法：如果在大字母前遇见小字母，那就将减去对应的小字母
代码实现
#include#include #include int romanToInt(char *s);int main(){ //今日分leetcode char *p = "LVIII"; printf("%d \n",romanToInt(p)); return 0;}/* 功能描述：将罗马字符串转成对应的数字*/int romanToInt(char *s){ int result = 0; while(*s) { switch (*s) { case 'I': if(*(s + 1) && *(s + 1) == 'V') { result += 4; s ++ ; } else if(*(s + 1) && *(s + 1) == 'X') { result += 9; s ++; } else { result += 1; } break; case 'V': result += 5; break; case 'X': if(*(s + 1) && *(s + 1) == 'L') { result += 40; s ++ ; } else if(*(s + 1) && *(s + 1) == 'C') { result += 90; s ++; } else { result += 10; } break; case 'L': result += 50; break; case 'C': if(*(s + 1) && *(s + 1) == 'D') { result += 400; s ++ ; } else if(*(s + 1) && *(s + 1) == 'M') { result += 900; s ++; } else { result += 100; } break; case 'D': result += 500; break; case 'M': result += 1000; break; default: break; } s ++; } return result;}
优化和改良
- 虽然Accepted，但是运行的时间和的所占内存还是有点大，重复的代码段有点多。
- 修改：将所有的重复代码部分提出来，放到最后统一执行
int romanToInt(char *s){ int left = 0,right = 0; int result = 0; while(*s) { left = ConvertToNumber(s); right = ConvertToNumber(s+1); if(left >= right) { result += left; } else { result = result + right - left; s ++; } s ++; } return result;}int ConvertToNumber(char *s){ switch (*s) { case 'I': return 1; case 'V': return 5; case 'X': return 10; case 'L': return 50; case 'C': return 100; case 'D': return 500; case 'M': return 1000; default: return 0; }} 
分析与总结
- 事实证明了，使用C去解决此类的算法问题是更快，但是写的确实要费劲一点！考虑要不要换一门语言？用Java？
转载地址：http://oggpb.baihongyu.com/