A message containing letters from A-Z is being encoded to numbers using the following mapping:

‘A’ -> 1
‘B’ -> 2
…
‘Z’ -> 26
Given an encoded message containing digits, determine the total number of ways to decode it.

For example,
Given encoded message “12”, it could be decoded as “AB” (1 2) or “L” (12).

The number of ways decoding “12” is 2.

這道題考察的是解碼，本質就是DFS深度優先遍歷。不過網上看了一個基於DP的解決方法，也很簡答。

程式碼如下：

import java.util.Arrays;

public class Solution
{
    int res=0;
    public int numDecodings(String s)
    {
        if(s==null || s.length()<=0)
            return res;

        //DP 做法
        int
 
 []nums=new int[s.length()+1];
        //num[0]必須初始化為1
        nums[0]=1;
        nums[1]=s.charAt(0)=='0'? 0:1;
        //下面是動態規劃去實現，
        for(int i=2;i<=s.length();i++)
        {
            if (s.charAt(i - 1) != '0')  
                nums[i] = nums[i - 1];  
            if (s.charAt(i - 2) != '0' && Integer.parseInt(""
 
 + s.charAt(i - 2) + s.charAt(i - 1)) < 27)  
                nums[i] += nums[i - 2];  
        } 
        return nums[s.length()];
    }

    private void getNum(String s, int index) 
    {
        if(index==s.length())
            res++;
        else
        {
            int a=(int)(s.charAt(index)-'0');
            if(a==0)
                return;

            getNum(s, index+1);
            if(index+1 < s.length() && a<=2 && a*10+(int)(s.charAt(index+1)-'0')<=26)
                getNum(s, index+2);
        }
    }
}

下面是C++的做法，就是一個DP動態規劃問題，這個和斐波那契額序列很像

程式碼如下：

#include <iostream>
#include <vector>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <queue>
#include <stack>
#include <string>
#include <climits>
#include <algorithm>
#include <sstream>
#include <functional>
#include <bitset>
#include <numeric>
#include <cmath>
#include <regex>
#include <iomanip>
#include <cstdlib>
#include <ctime>
using namespace std;

class Solution 
{
public:
    int numDecodings(string s) 
    {
        if (s.length() <= 0)
            return s.length();
        vector<int> dp(s.length() + 1, 0);
        dp[0] = 1;
        dp[1] = s[0]=='0'?0:1;
        for (int i = 2; i <= s.length(); i++)
        {
            if (s[i-1] != '0')
                dp[i] = dp[i - 1];
            if (s[i - 2] != '0' && stoi(s.substr(i-2,2))<=26)
                dp[i] += dp[i - 2];
        }
        return dp[s.length()];
    }
};