資訊專欄INFORMATION COLUMN
摘要：一個(gè)數(shù)字分為提取整十，提取再處理剩下小于的部分。大于的數(shù)字，有以內(nèi)的按上一行的方法處理。這樣窮舉出所有情況的好處，就是不用處理這種特殊情況，變得更迅速高效，代碼也簡潔。
Convert a non-negative integer to its english words representation.
123 -> "One Hundred Twenty Three" 12345 -> "Twelve Thousand Three Hundred Forty Five" 1234567 -> "One Million Two Hundred Thirty Four Thousand Five Hundred Sixty Seven"
邏輯本身不是太難。這題的特點(diǎn)就是英語是以1000為單位進(jìn)行操作的，不像漢語用“萬”。 英語的另一個(gè)特點(diǎn)就是小于20的數(shù)字要特殊對待，整十的數(shù)字也要特殊對待。 一個(gè)數(shù)字num, 分為num<20, num<100提取整十，num<1000提取hundreds,再處理剩下小于100的部分。 大于1000的數(shù)字，有thuansands, million, billion. 1000以內(nèi)的按上一行的方法處理。 這里另一個(gè)地方就是數(shù)字的單詞拼寫，不出錯(cuò)是很難的，所以多多記憶吧。
public class Solution {
private static final String[] lessThan20 = {"", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine",
"Ten", "Eleven", "Twelve", "Thirteen", "Fourteen","Fifteen","Sixteen","Seventeen","Eighteen","Nineteen"};
private static final String[] tens = {"","Ten","Twenty","Thirty","Forty","Fifty","Sixty","Seventy","Eighty","Ninety"};
private static final String[] thousands = {"","Thousand","Million","Billion"};
public String numberToWords(int num) {
if(num == 0) return "Zero";
String word = "";
int i = 0;
while(num > 0){
if(num%1000 != 0){
word = helper(num%1000) + thousands[i] + " " + word;
}
num = num/1000;
i++;
}
return word.trim();
}
public String helper(int n) {
if(n == 0){
return "";
} else if(n<20){
return lessThan20[n] + " ";
} else if(n<100){
return tens[n/10] + " " + helper(n%10);
} else {
return lessThan20[n/100] + " Hundred " + helper(n%100);
}
}
}
12 Integer to Roman
Given an integer, convert it to a roman numeral. Input is guaranteed to be within the range from 1 to 3999.
public class Solution {
public String intToRoman(int num) {
String[] M = {"","M","MM","MMM"}; // M=1000
String[] C = {"","C","CC","CCC","CD","D","DC","DCC","DCCC","CM"}; //C = 100, D= 500
String[] X = {"","X","XX","XXX","XL","L","LX","LXX","LXXX","XC"}; // X=10, L =50
String[] I = {"","I","II","III","IV","V","VI","VII","VIII","IX"}; // I= 1, V =5
return M[num/1000]+C[(num%1000)/100] + X[(num%100)/10] + I[(num%10)];
}
}
這樣窮舉出所有情況的好處，就是不用處理IV這種特殊情況，變得更迅速高效，代碼也簡潔。
相當(dāng)于簡化版的Integer to English Words.
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