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"
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 ret.substr(1, ret.length() - 1);
}
}
};
My code
class Solution {
public:
string digs [20]={"One","Two","Three","Four","Five","Six","Seven","Eight","Nine","Ten","Eleven","Twelve","Thirteen","Fourteen","Fifteen","Sixteen","Seventeen","Eighteen","Nineteen"};
string range [10]={"Twenty","Thirty","Forty","Fifty","Sixty","Seventy","Eighty","Ninety"};
string numberToWords(int num) {
if(num == 0){ return "Zero";}
else {
string ret = numString(num);
return ret.substr(0,ret.length()-1);}
}
string numString(int num){
if(num>=1000000000){
return numString(num/1000000000)+"Billion "+numString(num%1000000000);
}else if(num >= 1000000){
return numString(num/1000000)+"Million "+numString(num%1000000);
}else if(num >= 1000){
return numString(num/1000)+"Thousand "+numString(num%1000);
}else if(num >= 100){
return numString(num/100)+"Hundred "+numString(num%100);
}else if (num >= 20){
return range[((num/10)-2)]+" "+numString(num%10);
}else if (num >= 1){
return digs[num-1]+" ";
}else{
return "";
}
}
};
这道题让我们把一个整型数转为用英文单词描述，就像在check上写钱数的方法，我最开始的方法特别复杂，因为我用了几个switch语句来列出所有的单词，但是我看网上大神们的解法都是用数组来枚举的，特别的巧妙而且省地方，膜拜学习中。题目中给足了提示，首先告诉我们要3个一组的进行处理，而且题目中限定了输入数字范围为0到2^31 - 1之间，最高只能到billion位，3个一组也只需处理四组即可，那么我们需要些一个处理三个一组数字的函数，我们需要把1到19的英文单词都列出来，放到一个数组里，还要把20,30，... 到90的英文单词列出来放到另一个数组里，然后我们需要用写技巧，比如一个三位数n，百位数表示为n/100，后两位数一起表示为n%100，十位数表示为n%100/10，个位数表示为n%10，然后我们看后两位数是否小于20，小于的话直接从数组中取出单词，如果大于等于20的话，则分别将十位和个位数字的单词从两个数组中取出来。然后再来处理百位上的数字，还要记得加上Hundred。主函数中调用四次这个帮助函数，然后中间要插入"Thousand", "Million", "Billion"到对应的位置，最后check一下末尾是否有空格，把空格都删掉，返回的时候检查下输入是否为0，是的话要返回'Zero'。
class Solution {
public:
string numberToWords(int num) {
string res = convertHundred(num % 1000);
vector<string> v = {"Thousand", "Million", "Billion"};
for (int i = 0; i < 3; ++i) {
num /= 1000;
res = num % 1000 ? convertHundred(num % 1000) + " " + v[i] + " " + res : res;
}
while (res.back() == ' ') res.pop_back();
return res.empty() ? "Zero" : res;
}
string convertHundred(int num) {
vector<string> v1 = {"", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"};
vector<string> v2 = {"", "", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"};
string res;
int a = num / 100, b = num % 100, c = num % 10;
res = b < 20 ? v1[b] : v2[b / 10] + (c ? " " + v1[c] : "");
if (a > 0) res = v1[a] + " Hundred" + (b ? " " + res : "");
return res;
}
};
REF: