decode-ways

题目：

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 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.

程序：

方法一：

动态规划

class Solution {public:    int numDecodings(string s) {        if(s.empty())return 0;        vector<int> dp(s.size(),0);        dp[0]=s[0]-'0'>0?1:0;        for(int i=1;i<s.size();++i){            int tmp=stoi(s.substr(i-1,2));            dp[i]=(s[i]-'0'>0?dp[i-1]:0)+(tmp<=26 && tmp>=10?(i==1?1:dp[i-2]):0);        }        return dp[s.size()-1];    }};

方法二：

递归（此方法思想较容易理解，不过会内存超限）

class Solution {public:    int numDecodings(string s) {        set<string> res;        solve(res, s);        return res.size();    }    void solve(set<string> &res, string s)    {        if (s.size() == 0)        {            cout << str << endl;            res.insert(str);            //str.clear();            return;        }        for (int i = 0; i<1; i++)        {            if (s[i] == '0'&&i != s.size() - 1)            {                s = s.substr(i + 1);                continue;            }            if (s[i] == '0'&&i == s.size() - 1)            {                solve(res, string(""));            }            str.push_back(s[i] - '0' - 1 + 'A');            solve(res, s.substr(i + 1));            str.pop_back();            if (i + 1<s.size())            {                int val = (s[i]-'0') * 10 + (s[i + 1]-'0');                if (val<27)                {                    str.push_back(val - 1 + 'A');                    solve(res, s.substr(i + 2));                    str.pop_back();                }            }        }    }    string str;};