LeetCode 13. Roman to Integer

question:

复制代码
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.
复制代码

 

first:

复制代码
class Solution {
    public int romanToInt(String s) {
        char[] charArray = s.toCharArray();
        int result =0;
        for(int i=0;i<charArray.length;i++){
            if(charArray[i]=='I'&&(charArray[i+1]=='V'||charArray[i+1]=='X')){
                result--;
            }else if(charArray[i]=='X'&&(charArray[i+1]=='L'||charArray[i+1]=='C')){
                result -=10;
            }else if(charArray[i]=='C'&&(charArray[i+1]=='D'||charArray[i+1]=='M')){
                result -=100;
            }else{
                switch (charArray[i]){
                case 'I':result++;
                         break;
                case 'V':result+=5;
                         break;
                case 'X':result+=10;
                         break;
                case 'L':result+=50;
                         break;
                case 'C':result+=100;
                         break;
                case 'D':result+=500;
                         break;
                case 'M':result+=1000;
                }        
            }            
        }
        return result;
    }
}
复制代码

 

result:

复制代码
Run Code Status: Runtime Error
Run Code Result:
Your input

"III"

Your answer

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 3
    at Solution.romanToInt(Solution.java:6)
    at __DriverSolution__.__helper__(__Driver__.java:8)
    at __Driver__.main(__Driver__.java:52)
复制代码

second:

复制代码
class Solution {
    public int romanToInt(String s) {
        char[] charArray = s.toCharArray();
        int result =0;
        for(int i=0;i<charArray.length;i++){
            if(charArray[i]=='I'&&i+1<charArray.length&&(charArray[i+1]=='V'||charArray[i+1]=='X')){
                result--;
            }else if(charArray[i]=='X'&&i+1<charArray.length&&(charArray[i+1]=='L'||charArray[i+1]=='C')){
                result -=10;
            }else if(charArray[i]=='C'&&i+1<charArray.length&&(charArray[i+1]=='D'||charArray[i+1]=='M')){
                result -=100;
            }else{
                switch (charArray[i]){
                case 'I':result++;
                         break;
                case 'V':result+=5;
                         break;
                case 'X':result+=10;
                         break;
                case 'L':result+=50;
                         break;
                case 'C':result+=100;
                         break;
                case 'D':result+=500;
                         break;
                case 'M':result+=1000;
                }        
            }            
        }
        return result;
    }
}
复制代码

 

result:

 

conclusion:

 

posted @   Zhao_Gang  阅读(170)  评论(0编辑  收藏  举报
编辑推荐:
· Linux系列:如何用 C#调用 C方法造成内存泄露
· AI与.NET技术实操系列(二):开始使用ML.NET
· 记一次.NET内存居高不下排查解决与启示
· 探究高空视频全景AR技术的实现原理
· 理解Rust引用及其生命周期标识(上)
阅读排行:
· 阿里最新开源QwQ-32B,效果媲美deepseek-r1满血版,部署成本又又又降低了!
· 单线程的Redis速度为什么快?
· SQL Server 2025 AI相关能力初探
· 展开说说关于C#中ORM框架的用法!
· AI编程工具终极对决:字节Trae VS Cursor,谁才是开发者新宠?
点击右上角即可分享
微信分享提示