递归和动态规划解整数分解为字符串的问题

#include <iostream>
#include <vector> 
#include <unordered_map>
using namespace std;

class int2strcount{
public:
    static int DO(int num){
        if(num < 10) return 1;
        cout << "DO: " << num << endl;
        int target = num %100;
        if(target >9 && target <26)
          return DO(num/10) + DO(num/100);
        return DO(num/10);
    }
    static int DO2(int num){
        unordered_map<int, int> cache;
        return dfs(num, cache);
    }
    static int DO3(int num){
        if(num < 10) return 1;
        int len= 1;
        for(int i =num; i>10; i/=10) len++;
        vector<int> dp(len+1, 1);
        for(int i =2; i<=len; i++){
            int temp = num; // 使用临时变量处理num的修改
            for (int j = 0; j < len - i; j++) {
                temp /= 10; // 跳过前面的位数
            }
            cout<< "dp: " << temp << endl;
            int target = temp % 100;
            // cout<< "target: " << target << endl;
            if(target >9 && target <26)
                dp[i] = dp[i-1] + dp[i-2];
            else
                dp[i] = dp[i-1];
        }
        return dp[len];
    }
private:
    static int dfs(int num, unordered_map<int, int>& cache){
        if(cache.find(num) != cache.end()) return cache[num];
        if(num < 10) return 1;
        cout << "dfs: " << num << endl;
        int target = num %100;
        if(target >9 && target <26){
            cache[num] = dfs(num/10, cache) + dfs(num/100, cache);
          return cache[num];
        }
        cache[num] = dfs(num/10, cache);
        return cache[num];
    }
};
int main(){
    int2strcount ic;
    cout<<ic.DO(1255258)<<endl;
    cout<<ic.DO2(1255258)<<endl;
    cout<<ic.DO3(1255258)<<endl;
    return 0;
}

DO和DO1是递归的解法，DO1是DO的剪枝。DO3是动态规划的解法。

DO: 1255258
DO: 125525
DO: 12552
DO: 1255
DO: 125
DO: 12
DO: 1255
DO: 125
DO: 12
dg: 6
dfs: 1255258
dfs: 125525
dfs: 12552
dfs: 1255
dfs: 125
dfs: 12
dg: 6
dp: 12
dp: 125
dp: 1255
dp: 12552
dp: 125525
dp: 1255258
dp: 6