LeetCode (力扣) 13. Roman to Integer ( C ) - Easy- 暴力解

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此题为前一题 LeetCode 12. Integer to Roman 的延伸，这次为给定一个罗马数字转换成阿拉伯数字。

题目与范例如下

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, 2 is written as II in Roman numeral, just two one's added together. 12 is written as XII, which is simply X + II. The number 27 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.

Example 1:

Input: s = "III"
Output: 3
Example 2:

Input: s = "IV"
Output: 4
Example 3:

Input: s = "IX"
Output: 9
Example 4:

Input: s = "LVIII"
Output: 58
Explanation: L = 50, V= 5, III = 3.
Example 5:

Input: s = "MCMXCIV"
Output: 1994
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

Constraints:

1 <= s.length <= 15
s contains only the characters ('I', 'V', 'X', 'L', 'C', 'D', 'M').
It is guaranteed that s is a valid roman numeral in the range [1, 3999].

 

解题策略为依序拆解，先从前面的 'M' 转换成千，再转换 'CM' , 'CD' 900 和 400，再转换 'D'以此类推。

 

下方为我的代码

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

 

下方为时间与空间之消耗

Runtime: 4 ms, faster than 84.35 % of C online submissions for Roman to Integer.

Memory Usage: 5.8 MB, less than 58.45 % of C online submissions for Roman to Integer.