题目描述
罗马数字转整数,包含以下七种字符: I, V, X, L,C,D 和 M。分别对应1,5,10,50,100,500,1000。
- I 可以放在 V (5) 和 X (10) 的左边,来表示 4 和 9。
- X 可以放在 L (50) 和 C (100) 的左边,来表示 40 和 90。
- C 可以放在 D (500) 和 M (1000) 的左边,来表示 400 和 900。
给定一个罗马数字,将其转换成整数。输入确保在 1 到 3999 的范围内。
思路
将每一个对应罗马数字转换成数字,一个一个加起来。当小数字在大数字前时,需要减法。
eg:罗马数字IV,就是V-I,即5-1=4,其他正常叠加
难点
数字转换
方法
使用数组,给每一个罗马数字对应下标,每个下标在数组中对应值。
int romanToInt(char * s){
int num[26];// 优点在于利用ASCII码,数组空间尽可能小,也更能理解
num['I' - 'A'] = 1;
num['V' - 'A'] = 5;
num['X' - 'A'] = 10;
num['L' - 'A'] = 50;
num['C' - 'A'] = 100;
num['D' - 'A'] = 500;
num['M' - 'A'] = 1000;
int result = 0, len = strlen(s);
for(int i = 0; i < len; i++){
int value = num[s[i] - 'A'];
if(i < len-1 && value < num[s[i+1] - 'A']){
result -= value;
}else{
result += value;
}
}
return result;
}
java版本(题解中一个很好的思路)
该方法避免了在数字相加时进行判断,直接++
class Solution {
public int romanToInt(String s) {
// 优点在于将6种较为特殊的情况提前转化,在switch中进行判断即可,避免了小数字在大数字前的判断,且减少了switch语句判断的次数。
// 缺点在于当特例足够多时,会变得臃肿,不及+-判断精简
s = s.replace("IV","a");
s = s.replace("IX","b");
s = s.replace("XL","c");
s = s.replace("XC","d");
s = s.replace("CD","e");
s = s.replace("CM","f");
int result = 0;
for(int i = 0;i < s.length(); i++){
result += getValue(s.charAt(i));
}
return result;
}
private int getValue(char c) {
switch(c) {
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;
case 'a' : return 4;
case 'b' : return 9;
case 'c' : return 40;
case 'd' : return 90;
case 'e' : return 400;
case 'f' : return 900;
}
return 0;
}
}
吐槽一下个人代码问题
- 习惯于return 0,导致运行结果不符合预期,多return了一个0
- 大小写问题,V和v的问题,导致ASCII码变化,计算结果不符合预期
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最后编辑时间为: Sep 14, 2021 at 07:44 pm
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