《剑指offer》第四十六题：把数字翻译成字符串

// 面试题46：把数字翻译成字符串
// 题目：给定一个数字，我们按照如下规则把它翻译为字符串：0翻译成"a"，1翻
// 译成"b"，……，11翻译成"l"，……，25翻译成"z"。一个数字可能有多个翻译。例
// 如12258有5种不同的翻译，它们分别是"bccfi"、"bwfi"、"bczi"、"mcfi"和
// "mzi"。请编程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。

#include <string>
#include <iostream>

using namespace std;

int GetTranslationCount(const string& number);

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

    string numberInString = to_string(number);  //转为字符串
    return GetTranslationCount(numberInString);
}

// f(i) = f(i + 1) + g(i, i+1) * f(i + 2) 递归思想
int GetTranslationCount(const string& number)
{
    int length = number.length();
    int* counts = new int[length];  //记录每一位的可能翻译次数
    int count = 0;  //临时变量

    for (int i = length - 1; i >= 0; --i)  //从最后一位开始往前计算
    {
        count = 0;  //记得清零临时变量
        if (i < length - 1)  //计算 f(i + 1) 部分
            count = counts[i + 1];
        else
            count = 1;

        if (i < length - 1)  //计算 g(i, i+1) * f(i + 2) 部分
        {
            int num1 = number[i] - '0';
            int num2 = number[i + 1] - '0';
            int converted = num1 * 10 + num2;
            if (converted <= 25 && converted >= 10)  //g(i, i+1)
            {
                if (i < length - 2)
                    count += counts[i + 2];
                else
                    count += 1;
            }
        }
        counts[i] = count;  //更新当前位可能翻译数
    }
    count = counts[0];
    delete[] counts;

    return count;
}

// ====================测试代码====================
void Test(const string& testName, int number, int expected)
{
    if (GetTranslationCount(number) == expected)
        cout << testName << " passed." << endl;
    else
        cout << testName << " FAILED." << endl;
}

void Test1()
{
    int number = 0;
    int expected = 1;
    Test("Test1", number, expected);
}

void Test2()
{
    int number = 10;
    int expected = 2;
    Test("Test2", number, expected);
}

void Test3()
{
    int number = 125;
    int expected = 3;
    Test("Test3", number, expected);
}

void Test4()
{
    int number = 126;
    int expected = 2;
    Test("Test4", number, expected);
}

void Test5()
{
    int number = 426;
    int expected = 1;
    Test("Test5", number, expected);
}

void Test6()
{
    int number = 100;
    int expected = 2;
    Test("Test6", number, expected);
}

void Test7()
{
    int number = 101;
    int expected = 2;
    Test("Test7", number, expected);
}

void Test8()
{
    int number = 12258;
    int expected = 5;
    Test("Test8", number, expected);
}

void Test9()
{
    int number = -100;
    int expected = 0;
    Test("Test9", number, expected);
}

int main(int argc, char* argv[])
{
    Test1();
    Test2();
    Test3();
    Test4();
    Test5();
    Test6();
    Test7();
    Test8();
    Test9();

    return 0;
}

分析：有点像斐波那契数，不过加了个判断。