273. Integer to English Words

alighters 發布于2019-08-14 18:00 / 2231人閱讀

摘要：一個數字分為提取整十，提取再處理剩下小于的部分。大于的數字，有以內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。這題的特點就是英語是以1000為單位進行操作的，不像漢語用“萬”。
英語的另一個特點就是小于20的數字要特殊對待，整十的數字也要特殊對待。
一個數字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的數字，有thuansands, million, billion. 1000以內的按上一行的方法處理。

這里另一個地方就是數字的單詞拼寫，不出錯是很難的，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。
相當于簡化版的Integer to English Words.

云服務器 GPU云服務器 English Words Aspose.Words integer

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