91. Decode Ways

• 递归解法

  class Solution
  {
  public:
    int numDecodings(string s)
    {
        return getAns(s, 0);
    }
  private:
    int getAns(string s, int start)
    {
        if (start == s.length())
        {
            return 1;
        }

        if (s[start] == '0')
        {
            return 0;
        }

        int ans1 = getAns(s, start + 1);  //得到第一种划分的解码方式
        int ans2 = 0;

        if (start < s.length() - 1)
        {
            int ten = (int)(s[start] - '0') * 10;
            int one = (int)(s[start + 1] - '0') * 1;
            if (ten + one <= 26)
            {
                ans2 = getAns(s, start + 2);
            }
        }

        return ans1 + ans2;
    }
  };

这种解法超时。

• 带记忆的递归解法

    class Solution
    {
        public:
            int numDecodings(string s)
            {
                return getAns(s);
            }
        private:
            int getAns(string s)
            {
                if (s.length() == 0)
                {
                    return 1;
                }

                if (datamap.count(s))
                {
                    return datamap[s];  //s已经计算过 并存在了map中
                }
                if (s[0] == '0')
                {
                    return 0;
                }

                if (1 == s.length())
                {
                    return 1;
                }



                int ans1 = getAns(s.substr(1));  //得到第一种划分的解码方式

                const int prefix = stoi(s.substr(0, 2));
                int ans2 = 0;
                if (prefix <= 26)
                {
                    ans2 = getAns(s.substr(2));
                }

                datamap[s] = ans1 + ans2;

                return ans1 + ans2;
            }
        private:
            unordered_map<string, int> datamap;  //存储已经计算过的字符串
    };

• 由暴力递归而来的DP解法

class Solution
{
public:
    int numDecodings(string s)
    {
        int length = s.length();
        vector<int> dp(length + 1);
        int profix = 0;

        dp[length] = 1;

        if(s[length - 1] != '0')
        {
            dp[length - 1] = 1;
        }
        else
        {
            dp[length - 1] = 0;
        }


        for(int start = length - 2;start >= 0;start--)
        {
            if(s[start] == '0')
            {
                dp[start] = 0;
                continue;
            }

            dp[start] = dp[start + 1];

            profix = stoi(s.substr(start,2));

            if(profix <= 26)
            {
                dp[start] = dp[start] + dp[start + 2];
            }

        }

        return dp[0];
    }
};

• 进一步改进后的DP解法

  class Solution
  {
  public:
    int numDecodings(string s)
    {
        int length = s.length();
        int startnum = 0;  //dp[start]的值
        int start1 = 0;  //dp[start + 1]
        int start2 = 0;  //dp[start + 2]

        int profix = 0;

        start2 = 1;

        if (s[length - 1] != '0')
        {
            start1 = 1;
        }
        else
        {
            start1 = 0;
        }
        startnum = start1;

        for (int start = length - 2; start >= 0; start--)
        {
            if (s[start] == '0')
            {
                startnum = 0;
                start2 = start1;
                start1 = startnum;
                continue;
            }

            startnum = start1;

            profix = stoi(s.substr(start, 2));

            if (profix <= 26)
            {
                startnum = startnum + start2;
            }
            start2 = start1;
            start1 = startnum;
        }

        return startnum;
    }
  };

这次只需要线性空间的内存即可做到。

• 参考资料

  Decode Ways

  LeetCode 91. Decode Ways