[Leedcode 91] 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 "12" , it could be decoded as "AB" (1 2) or "L" (12).

The number of ways decoding is "12" 2.

 Public classSolution { Public intnumdecodings (String s) {/*The topic is similar to the number of steps in the question, so the need to use DP thought to save the middle of the sub-solution, but dp[i] is equal to dp[i-1] + dp[i-2] need to determine whether the character satisfies the requirements of the use of recursion will time out, so using the dynamic planning DP        method to do.        For string s[0...i] decoding number should be and s[0...i-1], S[0...i-2] The number of decoding is related. Dp[i]: The number of decoded on behalf of S[0...i-1], so the length of the DP is s length +1,dp[0]=dp[1]=1 dp[i] = {(s[i-1]!= ' 0 ')? dp[i-1]:0} + {s[i-2...i-1]<= ' 26 ' ?                Dp[i-2]: 0}; Sweep this string from start to finish, the string composed of s[0...i-1], how many kinds of decoding combinations, then there are two cases first: if S[I-1] corresponding to a single character can decode, then Dp[i] (S[0...i-1]) including the former dp[i-1]         The number of combinations accumulated dp[i] = dp[i-1] Second: If not only s[i-1] the corresponding single character can decode, S[i-2...i-1], two characters can also be decoded, then not only the dp[i-1] accumulated number of combinations, also including dp[i -2] bit accumulation dp[i] = Dp[i-1] + dp[i-2] We build an array of n+1, for the sake of brevity, we put a 1 in front of it. */        if(s==NULL|| S.length () <1)return0; if(S.charat (0) = = ' 0 ')return0; int[]dp=New int[S.length () +1]; dp[0]=1; Dp[1]=1;  for(intI=2;i<=s.length (); i++){            if(S.charat (i-1)! = ' 0 ') dp[i]=dp[i-1]; if(IsValid (S.substring (i-2,i))) dp[i]+=dp[i-2]; }        returndp[s.length ()]; }     Public BooleanIsValid (String s) {if(S.charat (0) = = ' 0 ')return false; if(S.charat (0) = = ' 1 ' | | (S.charat (0) = = ' 2 ' &&s.charat (1) <= ' 6 '))return true; return false; }}

[Leedcode 91] Decode Ways