LeetCode-13 罗马数字转整数(C)

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.

思路

这道题用C语言解的，就是if,else if套用，分各类情况，把IV``IX``XL``XC``CD``CM看成一个字符

解法1

字符串按倒序累加

int romanToInt(char* s) {
    int len=strlen(s);
    int result=0;
    for(int i=len-1;i>=0;i--)
    {
        if(s[i]=='I')
            result+=1;
        else if(s[i]=='V')
        {
            if(s[i-1]=='I')
            {
                result+=4;
                i--;
            }
            else
                result+=5;
        }
        else if(s[i]=='X')
        {
            if(s[i-1]=='I')
            {
                result+=9;
                i--;
            }
            else
                result+=10;
        }
        else if(s[i]=='L')
        {
            if(s[i-1]=='X')
            {
                result+=40;
                i--;
            }
            else 
                result+=50;
        }
        else if(s[i]=='C')
        {
            if(s[i-1]=='X')
            {
                result+=90;
                i--;
            }
            else
                result+=100;
        }
        else if(s[i]=='D')
        {
            if(s[i-1]=='C')
            {
                result+=400;
                i--;
            }
            else
                result+=500;
        }
        else if(s[i]=='M')
        {
            if(s[i-1]=='C')
            {
                result+=900;
                i--;
            }
            else
                result+=1000;
        }
    }
    return result;
}

解法2

字符串按正序累加

int romanToInt(char* s) {
    int i=0;
    int result=0;
    while(s[i]!='\0')
    {
        if(s[i]=='I')
        {
            if(s[i+1]=='V')
            {
                result+=4;
                i+=2;
            }
            else if(s[i+1]=='X')
            {
                result+=9;
                i+=2;
            }
            else 
            {
                result+=1;
                i++;
            }
        }
        else if(s[i]=='X')
        {
            if(s[i+1]=='L')
            {
                result+=40;
                i+=2;
            }
            else if(s[i+1]=='C')
            {
                result+=90;
                i+=2;
            }
            else 
            {
                result+=10;
                i++;
            }
        }
        else if(s[i]=='C')
        {
            if(s[i+1]=='D')
            {
                result+=400;
                i+=2;
            }
            else if(s[i+1]=='M')
            {
                result+=900;
                i+=2;
            }
            else 
            {
                result+=100;
                i++;
            }
        }
        else if(s[i]=='V')
        {
            result+=5;
            i++;
        }
        else if(s[i]=='L')
        {
            result+=50;
            i++;
        }
        else if(s[i]=='D')
        {
            result+=500;
            i++;
        }
        else if(s[i]=='M')
        {
            result+=1000;
            i++;
        }
    }
    return result;
}