每周算法：罗马数字转阿拉伯数字
Description
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.
Solution
Approach 1
这道题不是很难，很容易就能想到解法，需要注意好边界条件问题，下面就是我的解法。
pint romanToInt(string str) {
int ans = 0;
int pos = 0;
const int len = str.length();
// 千位
while (pos < len && str[pos] == 'M') {
++pos;
}
ans += 1000 * pos;
if (pos == len) {
return ans;
}
// 百位
switch (str[pos]) {
case 'C':
ans += 100;
++pos;
if (str[pos] == 'D') {
ans += 300;
++pos;
}
else if (str[pos] == 'M') {
ans += 800;
++pos;
}
else {
while (pos < len && str[pos] == 'C') {
ans += 100;
++pos;
}
}
break;
case 'D':
ans += 500;
++pos;
while (pos < len && str[pos] == 'C') {
ans += 100;
++pos;
}
break;
default:
break;
}
if (pos == len) {
return ans;
}
// 十位
switch (str[pos]) {
case 'X':
ans += 10;
++pos;
if (str[pos] == 'C') {
ans += 80;
++pos;
}
else if (str[pos] == 'L'){
ans += 30;
++pos;
}
else {
while (pos < len && str[pos] == 'X') {
ans += 10;
++pos;
}
}
break;
case 'L':
ans += 50;
++pos;
while (pos < len && str[pos] == 'X') {
ans += 10;
++pos;
}
break;
default:
break;
}
if (pos == len) {
return ans;
}
// 个位
switch (str[pos]) {
case 'I':
ans += 1;
++pos;
if (str[pos] == 'X') {
ans += 8;
++pos;
}
else if (str[pos] == 'V') {
ans += 3;
++pos;
}
else {
while (pos < len && str[pos] == 'I') {
++ans;
++pos;
}
}
break;
case 'V':
ans += 5;
++pos;
while (pos < len && str[pos] == 'I') {
++ans;
++pos;
}
break;
default:
break;
}
return ans;
}
不得不说，这样的代码比较丑，这是真的，就算解出来了成就感也不强。
Approach 2
这个解法是leetcode上最快的解法之一，只使用一个循环就完成了计算，代码整体上很简洁，也比较好理解。只需要针对6种特殊情况做特殊处理就好。
int romanToInt(string s) {
int result = 0;
int size = s.size();
for(int i = 0; i < size; ++i) {
switch(s[i]) {
case 'M':
if(i - 1 >= 0 && s[i - 1] == 'C')
result += 800;
else
result += 1000;
break;
case 'D':
if(i - 1 >= 0 && s[i - 1] == 'C')
result += 300;
else
result += 500;
break;
case 'C':
if(i - 1 >= 0 && s[i - 1] == 'X')
result += 80;
else
result += 100;
break;
case 'L':
if(i - 1 >= 0 && s[i - 1] == 'X')
result += 30;
else
result += 50;
break;
case 'X':
if(i - 1 >= 0 && s[i - 1] == 'I')
result += 8;
else
result += 10;
break;
case 'V':
if(i - 1 >= 0 && s[i - 1] == 'I')
result += 3;
else
result += 5;
break;
case 'I':
result += 1;
break;
default:;
}
}
return result;
}
Approach 3
下面这个是我在看该问题的discuss板块的时候发现的。不得不说，结题思路实在新奇，这点是值得学习的。不过我比较担心它的时间复杂度。该方法使用Java编写。
public int romanToInt(String s) {
int sum=0;
if(s.indexOf("IV")!=-1){sum-=2;}
if(s.indexOf("IX")!=-1){sum-=2;}
if(s.indexOf("XL")!=-1){sum-=20;}
if(s.indexOf("XC")!=-1){sum-=20;}
if(s.indexOf("CD")!=-1){sum-=200;}
if(s.indexOf("CM")!=-1){sum-=200;}
char c[]=s.toCharArray();
int count=0;
for(;count<=s.length()-1;count++){
if(c[count]=='M') sum+=1000;
if(c[count]=='D') sum+=500;
if(c[count]=='C') sum+=100;
if(c[count]=='L') sum+=50;
if(c[count]=='X') sum+=10;
if(c[count]=='V') sum+=5;
if(c[count]=='I') sum+=1;
}
return sum;
}