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[leetcode] Integer to English Words

2015-11-27 13:39 260 查看
题目:

Convert a non-negative integer to its english words representation. Given input is guaranteed to be less than 2^31 - 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"


分析:英语中只有十(ten),百(hundred),千(thousand),百万(million),十亿(billion),万亿(trillion),因为输入的数最大为2^31 - 1,故这里用不着万亿(trillion),其中1 trillion = 1000 billion = 10^6 million = 10^9 thousand = 10^12. 另外注意英语数字中小于20的单词都是不唯一不能组合的。

思路:从数字的低位开始,结合辅助数组,每三位(1000)处理一次,代码如下:

public class Solution {
String[] lessThan20 = {"", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"};
String[] tens = {"", "Ten", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"};
String hundred = "Hundred";
String[] thousands = {"", "Thousand", "Million", "Billion"};
public String numberToWords(int num) {
if (num == 0) {
return "Zero";
}
String result = "";
int i = 0;
while (num > 0) {
if (num % 1000 != 0) {
result = helper(num%1000) + thousands[i] + " " + result;
}
num /= 1000;
i++;
}
return result.trim(); //删除尾部的空格
}
private String helper(int i) {
if (i == 0) {
return "";
} else if (i < 20) {
return lessThan20[i] + " ";
} else if (i < 100) {
return tens[i/10] + " " + helper(i%10);
} else {
return helper(i/100) + hundred  + " " + helper(i%100);
}
}
}
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