[LC] Decode Ways

A message containing letters from was A-Z 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  "" ", it could be decoded as  "AB"   (1 2) Or  "L"   (a).

The number of ways decoding is "12" 2.

Class Solution {BOOL Judge (int i) {return i > 0 && i<3;        } bool Judge (int I,int j) {int t = i*10 + j;        if (t>26 | | t <=10) return false;    return true;        }public:int numdecodings (string s) {int n = s.size ();        int a[n];        for (int i = 0;i<n;++i) {A[i] = s[i]-' 0 ';        } if (n = = 0) return 0;        The first number is 0 if (a[0] = = 0) return 0;        if (n = = 1) return 1;        int dp[s.size ()];        Dp[0] = 1; if (a[1]! = 0) Dp[1] = Judge (a[0],a[1])?         2:1;            else{if (a[0]<=0 | | a[0] > 2) return 0;        else dp[1] = 1;                } if (n==2) return dp[1];  for (int i = 2; i<n; ++i) {if (a[i]! = 0) Dp[i] = Dp[i-1] + (judge (a[i-1],a[i))? Dp[i-2]:            0);          else if (a[i-1]<=0 | | a[i-1] > 2) return 0;      else dp[i] = dp[i-2];    } return dp[n-1]; }};

At the beginning did not consider the inside will appear 0, then is very silly than the submission of ...

can define DP[N],A[N]

The pseudo code is as follows,

If a[i] = = 0

If A[I-1] greater than 0 less than 3//such a[i-1] and a[i] can form a letter

Dp[i] = dp[i-2]//In this case, the number of species can only be equal to dp[i-2], the number of species does not increase

Else

Dp[i] = 0//illegal in other cases

Else

If a[i-1]<=0, greater than two

Dp[i] = dp[i-1]//New letters can only be one case

Else

Dp[i] = dp[i-1]+dp[i-2]//Add new letters as a case, you can also add new letters and old letters as a situation

Take a look at the Big Brother's answer

int numdecodings (string s) {    if (!s.size () | | | S.front () = = ' 0 ') return 0;    R2:decode ways of S[i-2], r1:decode ways of s[i-1]     int r1 = 1, r2 = 1;    for (int i = 1; i < s.size (), i++) {        //zero voids ways of the last because zero cannot is used separately        if ( S[i] = = ' 0 ') r1 = 0;        Possible two-digit letter, so new R1 are sum of both while new R2 are the old R1        if (s[i-1] = = ' 1 ' | | s[i-1] = = ' 2 ' && s[i] <= ' 6 ') {            r1 = r2 + r1;            r2 = r1-r2;        }        One-digit letter, no new to added        else {            r2 = r1;        }    }    return R1;}

That's a lot of information. The space is O (1), because only the first two values need to be stored.

[LC] Decode Ways