leetcode decode_ways

最新推荐文章于 2020-05-25 02:32:37 发布

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本文介绍了一种消息解码算法，该算法将字母映射为数字，并探讨了如何确定编码消息的解码方式总数。提供了递归、递归保存结果及动态规划三种实现方案。

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A message containing letters fromA-Zis being encoded to numbers using the following mapping:
‘A’ -> 1
‘B’ -> 2
…
‘Z’ -> 26
Given an encoded message containing , 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.

//递归
class Solution {
public:
    int numDecodings(string s) 
    {
        if (s.empty())
            return 0;
        for(int i = 0;i<s.size();i++)//存在异常时   如100 001  12505等
        {
            if (s[i] == '0')
            {
                if(i==0 || s[i-1] == '0' || s[i-1] > '2' )
                    return 0;
            }
        }
        
        vector<int> vec(s.size()+1, 0);
        return fun(s, s.size());
            
    }
    int fun(string &s, int n)
    {
        if(n<=1)
            return 1;
        if (s[n-1] == '0')
            return fun(s, n-2);
        else
            return ((s[n-2] == '1' ) || (s[n-2] == '2' && s[n-1] <= '6'))?fun(s, n-1) + fun(s, n-2):fun(s, n-1);
    }
};
            1                  n == 1|| 0
            f(n-2)            S[n] == '0'
f(n) =      f(n-2) + f(n-1)   s[n-1] == '1' || s[n-1] == '2' && s[n] <= '6'
            f(n-1)             else

//递归 保存结果
class Solution {
public:
    int numDecodings(string s) 
    {
        if (s.empty())
            return 0;
        for(int i = 0;i<s.size();i++)//存在异常时   如100 001  12505等
        {
            if (s[i] == '0')
            {
                if(i==0 || s[i-1] == '0' || s[i-1] > '2' )
                    return 0;
            }
        }
        
        vector<int> vec(s.size()+1, 0);
        return fun(s, s.size(), vec);
            
    }
    int fun(string &s, int n, vector<int> &vec)
    {
        if(n<=1)
            return 1;
        if (vec[n] != 0) 
            return vec[n];
        if (s[n-1] == '0')
            vec[n] = fun(s, n-2,vec);
        else
            vec[n] = ((s[n-2] == '1' ) || (s[n-2] == '2' && s[n-1] <= '6'))?fun(s, n-1,vec) + fun(s, n-2,vec):fun(s, n-1,vec);
        return vec[n] ;
    }
};

//动态规划

class Solution {
public:
    int numDecodings(string s) 
    {
        if (s.empty())
            return 0;
        for(int i = 0;i<s.size();i++)//存在异常时   如100 001  12505等
            if (s[i] == '0')
                if(i==0 || s[i-1] == '0' || s[i-1] > '2' )
                    return 0;
        vector<int> vec(s.size()+1, 0);
        for (int i=0;i<=s.size();i++)
        {
            if(i<=1)
                vec[i] = 1;
            else if (s[i-1] == '0')
                vec[i] = vec[i-2];
            else
                vec[i] = ((s[i-2] == '1' ) || (s[i-2] == '2' && s[i-1] <= '6'))?vec[i-1] + vec[i-2]:vec[i-1];
        }
        return vec[s.size()];
    }
};

测试

#include"head.h"
class Solution {
public:
    int numDecodings(string s) 
    {
        if (s.empty())
            return 0;
        for(int i = 0;i<s.size();i++)//存在异常时   如100 001  12505等
        {
            if (s[i] == '0')
            {
                if(i==0 || s[i-1] == '0' || s[i-1] > '2' )
                    return 0;
            }
        }
        
        vector<int> vec(s.size()+1, 0);
        return fun(s, s.size());
            
    }
    int fun(string &s, int n)
    {
        if(n<=1)
            return 1;
        if (s[n-1] == '0')
            return fun(s, n-2);
        else
            return ((s[n-2] == '1' ) || (s[n-2] == '2' && s[n-1] <= '6'))?fun(s, n-1) + fun(s, n-2):fun(s, n-1);
    }
};


//递归 保存
class Solution1 {
public:
    int numDecodings(string s) 
    {
        if (s.empty())
            return 0;
        for(int i = 0;i<s.size();i++)//存在异常时   如100 001  12505等
        {
            if (s[i] == '0')
            {
                if(i==0 || s[i-1] == '0' || s[i-1] > '2' )
                    return 0;
            }
        }
        
        vector<int> vec(s.size()+1, 0);
        return fun(s, s.size(), vec);
            
    }
    int fun(string &s, int n, vector<int> &vec)
    {
        if(n<=1)
            return 1;
        if (vec[n] != 0) 
            return vec[n];
        if (s[n-1] == '0')
            vec[n] = fun(s, n-2,vec);
        else
            vec[n] = ((s[n-2] == '1' ) || (s[n-2] == '2' && s[n-1] <= '6'))?fun(s, n-1,vec) + fun(s, n-2,vec):fun(s, n-1,vec);
        return vec[n] ;
    }
};

class Solution2 {
public:
    int numDecodings(string s) 
    {
        if (s.empty())
            return 0;
        for(int i = 0;i<s.size();i++)//存在异常时   如100 001  12505等
            if (s[i] == '0')
                if(i==0 || s[i-1] == '0' || s[i-1] > '2' )
                    return 0;
        vector<int> vec(s.size()+1, 0);
        for (int i=0;i<=s.size();i++)
        {
            if(i<=1)
                vec[i] = 1;
            else if (s[i-1] == '0')
                vec[i] = vec[i-2];
            else
                vec[i] = ((s[i-2] == '1' ) || (s[i-2] == '2' && s[i-1] <= '6'))?vec[i-1] + vec[i-2]:vec[i-1];
        }
        return vec[s.size()];
    }
};
int main()
{
    string str = "11";
    Solution s;
    cout<<"递归"<<s.numDecodings(str)<<endl;
    
    Solution1 s1;
    cout<<"递归保存结果"<<s1.numDecodings(str)<<endl;
    
    Solution2 s2;
    cout<<"动态规划结果"<<s2.numDecodings(str)<<endl;
}