Leetcode——91. Decode Ways

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https://leetcode.com/problems/decode-ways/?tab=Solutions
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.

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动态规划问题：递推公式为

DP问题：dp[i+2]=s[i,i+1]<=26？2*dp[i]:dp[i]

dp[i]表示到i位置，可能的解码顺序。

可是这一题需要考虑300,00，这种有0出现的情况，这些边边角角真TM烦人！
本来十来行代码，现在为了检查意外情况一下子变成几十行了！

class Solution {//DP问题：dp[i+2]=(s[i,i+1]<=26？2*dp[i]:dp[i])public:    int numDecodings(string s) {        if(s.length()==0) return 0;        int *dp=new int[s.length()];        int *num=new int[s.length()];        for(int i=0;i<s.length();i++)        {            num[i]=s[i]-'0';        }        //合理性检查，排除30,00这种数的出现        if(num[0]==0)            return 0;        else            for(int i=1;i<s.length();i++)            {                if((num[i]==0&&num[i-1]>=3)||(num[i]==0&&num[i-1]==0)) return 0;                else continue;            }        dp[0]=1;        if(s.length()>=2)        {            int temp=num[0]*10+num[1];            if(temp<=26&&num[1]!=0)                dp[1]=2;            else                dp[1]=1;        }        for(int i=2;i<s.length();i++)        {            if(num[i-1]==0)//前面的判断已经保证不会出现00                dp[i]=dp[i-1];            else if(num[i]==0)                {                    if(num[i-1]<3)//出现20这种                        dp[i]=dp[i-2];                    else                        return 0;                }            else            {                 int temp=num[i-1]*10+num[i];                if(temp<=26)                    dp[i]=dp[i-1]+dp[i-2];                else                    dp[i]=dp[i-1];            }        }        return dp[s.length()-1];    }};