[LeetCode] 273. Integer to English Words

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整数转换英文表示。题意是给一个整数，请转换成英文表示。例子，

Example 1:

Input: 123
Output: "One Hundred Twenty Three"

Example 2:

Input: 12345
Output: "Twelve Thousand Three Hundred Forty Five"

Example 3:

Input: 1234567
Output: "One Million Two Hundred Thirty Four Thousand Five Hundred Sixty Seven"

Example 4:

Input: 1234567891
Output: "One Billion Two Hundred Thirty Four Million Five Hundred Sixty Seven Thousand Eight Hundred Ninety One"

这个题目不涉及任何算法，是一道单纯的实现题。需要设计一个helper函数，判断如下几个条件

num是否等于0

num是否小于20

num是否小于100

num % 1000是否等于0 - 为什么是%1000是因为input最大只会有10个digit，最大的数字只会是到billion，是由1000表达法来的，这也是英文数字的特点，每三位就一定会有一个单词（hundred, thousand, million, billion）

细节方面，首先需要创建几个string array，以方便输出一些单词。接着判断这个数字num % 1000是否为0，若不为0则说明这个数字大于1000，则输出里面一定包含thousand那一组里面的词，接着再除以1000，递归处理之后的数字；如果num % 1000 == 0，则表示要处理的部分小于1000，这里又分另外几种情况，

如果数字小于20，直接在less20里面找对应的数字

如果20 <= num < 100，在tens里面找num / 10然后helper递归找num % 10

如果num >= 100，在less20里面找一个单词跟hundred拼接，再递归找num % 100

跑一个例子，12345 -> "Twelve Thousand Three Hundred Forty Five"

首先在line13，12345 % 1000 = 345 != 0，则res = helper(num % 1000) + thousands[i] + " " + res; => res = helper(345) + "" + " " + res;

此时num = num / 1000 = 12, i = 1；此时掉入helper函数的递归，此时又因为num = 345所以是会掉入line 31的if语句，所以会在less20里面找到less20[345 / 100] = less20[3] = Three + " Hundred " + helper(345 % 100) => "Three Hundred " + helper(45) => "Three Hundred " + tens[45 / 10] + " " + helper(45 % 10) => "Three Hundred " + "Forty" + " " + helper(5) => "Three Hundred " + "Forty" + " " + "Five"。

完成了345的部分之后，helper函数会回溯，此时num = num / 1000 = 12, i = 1的关系所以helper函数判断会掉入27行得到前面的"Twelve Thousand"，再 + res得到最后的结果"Twelve Thousand Three Hundred Forty Five"。

时间O(n)

空间O(1) - 只有3个额外的数组

Java实现

 1 class Solution {
 2     String[] less20 = { "", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten", "Eleven",
 3             "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen" };
 4     String[] tens = { "", "Ten", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety" };
 5     String[] thousands = { "", "Thousand", "Million", "Billion" };
 6 
 7     public String numberToWords(int num) {
 8         // corner case
 9         if (num == 0)
10             return "Zero";
11         String res = "";
12         int i = 0;
13         while (num > 0) {
14             if (num % 1000 != 0) {
15                 res = helper(num % 1000) + thousands[i] + " " + res;
16             }
17             num /= 1000;
18             i++;
19         }
20         return res.trim();
21     }
22 
23     public String helper(int num) {
24         if (num == 0)
25             return "";
26         if (num < 20) {
27             return less20[num % 20] + " ";
28         } else if (num < 100) {
29             return tens[num / 10] + " " + helper(num % 10);
30         } else {
31             return less20[num / 100] + " Hundred " + helper(num % 100);
32         }
33     }
34 }

[LeetCode] 273. Integer to English Words

标签：cas   实现   拼接   san   一个   单词   空间   数字   设计   

原文地址：https://www.cnblogs.com/aaronliu1991/p/12522025.html