Decode Ways——动态规划

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.

这题使用动态规划来做，但是刚开始的时候一直超时，我以为是其他地方的原因，但是怎么改都有问题（代码如下）后来，我发现我这样做压根就不是动态规划啊，就是普通的递归，怪不得会超时·················动态规划与普通的递归的区别就是在于不要重复计算啊！！！！！动态规划得到初始的0,1，然后由0,1计算得到后面的数·······

class Solution {
public:
int numDecodings(string s) {
int len=s.size();
if(len==0||s[0]=='0')
return 0;
int num;
int res;
if(len==1)
num =s[0]-'0';
else if(len==2)
num=10*(s[0]-'0')+s[1]-'0';
if(num==0)
return 0;
else if(num<=10)
return 1;
else if(num>10&&num<=26)
return 2;
else if(num>26&&num<=99)
return 1;
res=max(numDecodings(s.substr(0,1))+numDecodings(s.substr(1,len)),
numDecodings(s.substr(0,2))+numDecodings(s.substr(2,len)));
return res;
}
};

正确的做法应该是这样的：

class Solution {
public:
int numDecodings(string s) {
int len=s.size();
if(len==0)
return 0;
else if(len==1)
return s[0]!='0'?1:0;
else if(len==2)
return (s[0]!='0'&&s[1]!='0'?1:0)+(s[0]!='0'&&((10*(s[0]-'0')+s[1]-'0')<=26)?1:0);

int* flag=new int[len];
flag[0]=s[0]!='0'?1:0;
flag[1]=(s[0]!='0'&&s[1]!='0'?1:0)+(s[0]!='0'&&((10*(s[0]-'0')+s[1]-'0')<=26)?1:0);
for(int i=2;i<len;i++)
{
flag[i]=0;
if(s[i]!='0')
{
flag[i]+=flag[i-1];
}
if(s[i-1]!='0'&&(10*(s[i-1]-'0')+s[i]-'0'<=26))
{
flag[i]+=flag[i-2];
}
}
int res=flag[len-1];
delete []flag;
return res;
}
};