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  • [LeetCode] 273. Integer to English Words

    Convert a non-negative integer num to its English words representation.

    Example 1:

    Input: num = 123
    Output: "One Hundred Twenty Three"
    

    Example 2:

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

    Example 3:

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

    Example 4:

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

    Constraints:

    • 0 <= num <= 231 - 1

    整数转换英文表示。

    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,这是因为英语里面数字的读法都是每三位一个单词,所以每次%1000就会知道到底是thousand、million还是billion(这是伴随着i++而记录的)。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;num被处理到小于1000之后,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,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,此时因为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 题目总结

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