[sword finger offer] 46 Translate numbers into strings

The finger of the sword Offer 46. Translate numbers into strings

subject

Given a number , We translate it as a string according to the following rules ：0 Translate into “a” ,1 Translate into “b”,……,11 Translate into “l”,……,25 Translate into “z”. A number may have more than one translation . Please program a function , Used to calculate how many different translation methods a number has .

Ideas

• Definition f[i] Before presentation i How many different translation methods are there for a number . that ,f[n] Just before n How many different translation methods are there for a number .
• Separate translation s[i]. Because we are looking for the number of schemes , If the second... Is determined i The translation of numbers , So before translation i A number and before translation i - 1 The number of methods is the same , namely f[i] = f[i - 1].
• take s[i] and s[i - 1] Combine to translate （ The combined numbers range from 10~25）. If the second... Is determined i The first number and the second number i - 1 The translation of numbers , So before translation i A number and before translation i-2 The number of translation methods for numbers is the same , namely f[i] = f[i - 2];
• Finally, add up the number of schemes of the two decisions , So the state transition equation is ：f[i] = f[i - 1] + f[i - 2];

Code

class Solution {
    
public:
    int translateNum(int num) {
    
        string s = to_string(num);//  Convert numbers to strings 
        int n = s.size();

        vector<int> f(n + 1);

        f[0] = 1;//  initialization 


        for(int i = 1; i <= n; i++)
        {
    
            f[i] = f[i - 1];//  Separate translation s[i]

            if(i > 1)
            {
    
                int t = (s[i - 2] - '0') * 10 + s[i - 1] - '0';
                if(t >= 10 && t <= 25)
                {
    
                    f[i] += f[i - 2];//  If you qualify   Direct will s[i] and s[i - 1] Combinatorial translation 
                }
            }
        }
        return f[n];
    }
};