leetcode 273. Integer to English Words

àì夳堔傛蜴生んèń 2022-07-27 14:52 126阅读 0赞

Convert a non-negative integer to its english words representation. Given input is guaranteed to be less than 231 - 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"

这题纯粹考细心，其他没什么说的

const string onedigit[9] = { "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine" };
    const string teensless20[10] = { "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen" };
    const string teens[8] = { "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety" };
    
    class Solution {
    	string get_digit(int num)
    	{
    		return onedigit[num-1];
    	}
    	string get_teens(int num)
    	{
    		if (num < 20)
    			return teensless20[num - 10];
    		return teens[num / 10 - 2];
    	}
    	string get_hundred(int num)
    	{
    		string re;
    		if (num >= 100)
    			re += get_digit(num / 100) + " Hundred ";
    		if (num - num / 100 * 100>=10)
    			re +=get_teens(num - num / 100 * 100)+" ";
    		if (num % 10 != 0 && (num - num / 100 * 100 >= 20 || num - num / 100 * 100<10))
    			re += get_digit(num % 10);
    		while (re.back() == ' ')
    			re.pop_back();
    		return re;
    	}
    public:
    	string numberToWords(int num) {//0-2 147 483 647
    		if (num == 0)
    			return "Zero";
    		string re;
    		vector<int>digits;
    		while (num != 0)
    		{
    			digits.push_back(num % 10);
    			num /= 10;
    		}
    		if (digits.size() > 9)
    			re += get_digit(digits[9])+" Billion ";
    		if (digits.size() > 6)
    		{
    			int jmax = digits.size() >= 9 ? 8 : digits.size()-1;
    			int sum = 0;
    			for (int j = jmax; j >= 6; j--)
    				sum = 10 * sum + digits[j];
    			if (sum>0)
    			re += get_hundred(sum)+" Million ";
    		}
    		if (digits.size() > 3)
    		{
    			int jmax = digits.size() >= 6 ? 5 : digits.size()-1;
    			int sum = 0;
    			for (int j = jmax; j >= 3; j--)
    				sum = 10 * sum + digits[j];
    			if (sum>0)
    			re += get_hundred(sum) + " Thousand ";
    		}
    		int jmax = digits.size() >= 3 ? 2 : digits.size()-1;
    		int sum = 0;
    		for (int j = jmax; j >= 0; j--)
    			sum = 10 * sum + digits[j];
    		if (sum>0)
    		re += get_hundred(sum);
    		while (re.back() == ' ')
    			re.pop_back();
    		return re;
    	}
    };

accepted