Leetcode #91 Decode Ways

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.

Difficulty:Medium

解法一，NP法，超时了。

int count(string s){
int ans = 0;
int len = s.length();
if(len==0)
return 1;
if(len==1)
return 1;
if(len==2&&s[0]-'0'>=2&&s[1]-'0'>6)
return 1;
if(len==2&&s[0]-'0'<=2&&s[1]-'0'<=6)
return 2;
//cout<<"111"<<endl;
string s1,s2,s3;
s1.push_back(s[0]);
for(int i = 1;i<len;i++)
s2.push_back(s[i]);
ans = ans + count(s1)*count(s2);
//cout<<ans<<endl;
s1.push_back(s[1]);
if(s1[0]-'0'<=2&&s1[1]-'0'<=6)
{
for(int i = 2;i<len;i++)
s3.push_back(s[i]);
ans = ans + count(s3);
}
return ans;
}
int numDecodings(string s) {
if(s.length()==0)
return 0;
return count(s);
}

解法二：用一个数组来记录前i（0-n）个的可能性，然后根据条件判断是nums[i]=nums[i-1]还是nums[i]=nums[i-1]+nums[i-2]，条件比较绕，WA了好多次。

int numDecodings(string s) {
int len = s.length();
vector<int> nums(len+1,0);
if(len==0)
return 0;
if(s[0]=='0')
return 0;
if(len==1)
return 1;
nums[0] = 1;nums[1] = 1;

for(int i = 2;i<=len;i++)
{
if(s[i-1]!='0')
nums[i] = nums[i-1];
if(s[i-2]!='0')
if((s[i-2]-'0'==1)||(s[i-2]-'0'==2&&s[i-1]-'0'<=6))
nums[i] = nums[i]+nums[i-2];
}
return nums[len];
}