46 把数字翻译成字符串

题目

给定一个数字，按照如下规则翻译成字符串：0翻译成“a”，1翻译成“b”...25翻译成“z”。一个数字有多种翻译可能，例如12258一共有5种，分别是bccfi，bwfi，bczi，mcfi，mzi。实现一个函数，用来计算一个数字有多少种不同的翻译方法。

C++题解

以12258为例,从最大的问题开始，递归 ：

递推公式： f(r-2) = f(r-1)+g(r-2,r-1)*f(r)
其中，如果r-2，r-1能够翻译成字符，则g(r-2,r-1)=1，否则为0。
因此，对于12258：

f(5) = 0
f(4) = 1
f(3) = f(4)+0 = 1
f(2) = f(3)+f(4) = 2
f(1) = f(2)+f(3) = 3 
f(0) = f(1)+f(2) = 5

int GetTranslationCount(const string& number);

int GetTranslationCount(int number)
{
	if (number < 0)
		return 0;

	string numberInString = to_string(number);
	return GetTranslationCount(numberInString);
}

int GetTranslationCount(const string& number)
{
	int length = number.length();
	int* counts = new int[length];
	int count = 0;

	for (int i = length - 1; i >= 0; --i)
	{
		count = 0;
		if (i < length - 1)
			count = counts[i + 1];
		else
			count = 1;

		if (i < length - 1)
		{
			//求出临近的两位数
			int digit1 = number[i] - '0';
			int digit2 = number[i + 1] - '0';
			int converted = digit1 * 10 + digit2;

			if (converted >= 10 && converted <= 25)
			{
				if (i < length - 2)
					count += counts[i + 2];
				else
					count += 1;
			}
		}

		counts[i] = count;
	}

	count = counts[0];
	delete[] counts;

	return count;
}

python 题解

class Solution:
    def getTranslationCount(self, number):
        """
        :type number: int
        :rtype: int
        """
        if number<0:
            return 0
        numberStr=str(number)

        return self.getTranslateCount(numberStr)
    
    def getTranslateCount(self,numberStr):
        length=len(numberStr)
        counts=[0]*length
        #count=0
        for i in range(length-1,-1,-1):
            count=0
            if i <length-1:
                count+=counts[i+1]
            else:
                count=1
            
            if i<length-1:
                digit1=int(numberStr[i])
                digit2=int(numberStr[i+1])
                converted=digit1*10+digit2
                if converted>=10 and converted<=25:
                    if i<length-2:
                        count+=counts[i+2]
                    else:
                        count+=1
            counts[i]=count
        return counts[0]