LeetCode刷题2

in 力扣 with 121 comments

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。

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

  1. 习惯于return 0,导致运行结果不符合预期,多return了一个0
  2. 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
Responses
  1. |Use the entire beauty product up before throwing them out. You just have to get the most out of what you buy. Try to get the last of the product by turning them upside down and squeezing the last bit out. Another tip is to remove the top of the bottle so you can reach into the bottle to get any remnants. You can wind up saving a lot of money by doing this.

    Reply