最近刷了很多leetcode, 无聊的题目就不写了,写点有意思的,比如这道阿拉伯数字转罗马数字。
题目给了几个例子,具体来说就是每一位的49和其他不一样,其他都是累加。
那么根据每一位的不同,我们按位计算,代码虽然多了一点,但是能解决问题。每一位最多判断四次。
package com.Int_to_Roman;
import java.util.Collections;
public class Solution {
public String Ge(int num) {
if (num<4)
return String.join("", Collections.nCopies(num, "I"));
else if (num == 4)
return "IV";
else if (num == 9)
return "IV";
else
return "V"+String.join("", Collections.nCopies(num-5, "I"));
}
public String Shi(int num) {
if (num<4)
return String.join("", Collections.nCopies(num, "X"));
else if (num == 4)
return "XL";
else if (num == 9)
return "XC";
else
return "L"+String.join("", Collections.nCopies(num-5, "X"));
}
public String Bai(int num) {
if (num<4)
return String.join("", Collections.nCopies(num, "C"));
else if (num == 4)
return "CD";
else if (num == 9)
return "CM";
else
return "D"+String.join("", Collections.nCopies(num-5, "C"));
}
public String Qian(int num) {
return String.join("", Collections.nCopies(num, "M"));
}
public String intToRoman(int num) {
String res = "";
int count = 0;
while(num !=0) {
int mod = num%10;
num = num/10;
if (count==0)
res = Ge(mod)+res;
else if (count == 1)
res = Shi(mod)+res;
else if (count == 2)
res = Bai(mod)+res;
else
res = Qian(mod)+res;
count+=1;
}
return res;
}
public static void main(String[] args) {
// TODO Auto-generated method stub
Solution s = new Solution();
System.out.println(s.intToRoman(1994));
}
}
提交虽然成功了,但是看到有位大神写得更加简单的方法,就是每次从大到小,减去相应的数字,再添加上这些数字对应的罗马数字。简单到没朋友,佩服佩服,而我就只会用这种近乎蠢的办法。还是要学习一下。
public class Solution {
public String intToRoman(int num) {
int[] values = {1000,900,500,400,100,90,50,40,10,9,5,4,1};
String[] strs = {"M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"};
StringBuilder sb = new StringBuilder();
for(int i=0;i<values.length;i++) {
while(num >= values[i]) {
num -= values[i];
sb.append(strs[i]);
}
}
return sb.toString();
}