问题
Convert a non-negative integer to its english words representation. Given input is guaranteed to be less than 231 - 1.
例子
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
int的最大值为2147483647,所以一个int最多可以被表示成X Billion X Million X Thousand X,其中X部分又可以被表示成Y Hundred Z. 所以解决问题的关键是把数字切割成若干部分,每一部分都可以被简单地表示出来。
要点
- 映射表
- 分类讨论
时间复杂度
O(n), n为数字位数
空间复杂度
O(1)
代码
class Solution {
public:
string numberToWords(int num) {
if (num == 0) return string("Zero");
string res;
for (int i = 3; i >= 0; i--) {
int div = pow(1000, i);
int n = (num - num % div) / div;
if (n != 0) {
res += toStr000(n) + " ";
if (num >= 1000)
res += str00[i] + " ";
}
num %= div;
}
res.pop_back();
return res;
}
private:
string toStr000(int num) {
string res;
int n = num / 100;
if (n != 0)
res += str1[n] + " " + str00[0] + " ";
num %= 100;
n = num / 10;
int d = num % 10;
if (n != 0) {
if (n == 1) res += str1x[d] + " ";
else res += strx0[n - 2] + " ";
}
if (n != 1 && d != 0)
res += str1[d] + " ";
res.pop_back();
return res;
}
vector<string> str1{ "Zero", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine" };
vector<string> str1x{ "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen" };
vector<string> strx0{ "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety" };
vector<string> str00{ "Hundred", "Thousand", "Million", "Billion" };
};
more concise
class Solution {
public:
string digits[20] = {"Zero", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"};
string tens[10] = {"Zero", "Ten", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"};
string int2string(int n) {
if (n >= 1000000000) {
return int2string(n / 1000000000) + " Billion" + int2string(n % 1000000000);
} else if (n >= 1000000) {
return int2string(n / 1000000) + " Million" + int2string(n % 1000000);
} else if (n >= 1000) {
return int2string(n / 1000) + " Thousand" + int2string(n % 1000);
} else if (n >= 100) {
return int2string(n / 100) + " Hundred" + int2string(n % 100);
} else if (n >= 20) {
return " " + tens[n / 10] + int2string(n % 10);
} else if (n >= 1) {
return " " + digits[n];
} else {
return "";
}
}
string numberToWords(int num) {
if (num == 0) {
return "Zero";
} else {
string ret = int2string(num);
return std::move(ret.begin()+1, ret.end(), std::back_inserter(result));
}
}
};
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