[LeetCode 273] Integer to English Word

本文介绍了一种将非负整数转换为其英文单词表示的方法。通过将数字按千位分组,并针对每三位数字编写辅助函数来实现。讨论了多种特殊情况处理方式,包括0和中间存在零的情况。

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Convert a non-negative integer to its english words representation. Given input is guaranteed to be less than 231 - 1.

For example,

123 -> "One Hundred Twenty Three"
12345 -> "Twelve Thousand Three Hundred Forty Five"
1234567 -> "One Million Two Hundred Thirty Four Thousand Five Hundred Sixty Seven"

Hint:

  1. Did you see a pattern in dividing the number into chunk of words? For example, 123 and 123000.
  2. Group the number by thousands (3 digits). You can write a helper function that takes a number less than 1000 and convert just that chunk to words.
  3. There are many edge cases. What are some good test cases? Does your code work with input such as 0? Or 1000010? (middle chunk is zero and should not be printed out)
solution:

Just as hint described, handle 3digit at each time, convert them into English word,

every time handle 3 digit, the unit will increase.

So many corner cases need to handle.

public String numberToWords(int num) {
        String[] units = {""," Thousand"," Million"," Billion"};
        int i = 0;
        String res="";
        while(num>0) {
            int temp = num%1000;
            if(temp>0) res = convert(temp) + units[i] + (res.length()==0 ?"": " "+res);
            num /= 1000;
            i++;
        }
        return res.isEmpty()? "Zero" : res;
    }
    public String convert(int num){
        String res = "";
        String[] ten = {"Zero", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine"};
        String[] hundred = {"Ten", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"};
        String[] twenty = {"Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"};
        if(num>0) {
            int temp = num/100;
            if(temp>0) {
                res += ten[temp] + " Hundred";
            }
            temp = num%100;
            if(temp>=10 && temp<20){
                if(!res.isEmpty()) res +=" ";
                res = res + twenty[temp%10];
                return res;
            }else if(temp>=20){
                temp = temp/10;
                if(!res.isEmpty()) res +=" ";
                res = res + hundred[temp-1];
            }
            temp = num%10;
            if(temp>0) {
                if(!res.isEmpty()) res +=" ";
                res = res + ten[temp];
            }
        }
        return res;
    }

 



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