[leetcode#91] Decode Ways

problem:

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.

Analysis:

Note the transitional function!!!at first glance, it could is solved through recursive call. But we are about the total count!!!Why we must follow that routinue. The difference here is only:count the present digit as a character, or count both digits as a character. For all chacter, we should ask us the same question. Since there is only a possible situations, why not we use dynamic programming forit. The range of digits in ThisProblem is ' 00 '-' 99 ', we could intepret those, digits separately or together. Suppose Res[i] represents the sequences count from0To I.1. ' 00 'Res[i]= 0, since ' xx ' is invalid and ' 0 'is invalid.2. ' 01 '-' 09 'Res[i]= Res[i-1], must be sperated.3. ' 10 ', ' 20 'Res[i]= Res[i-2], theNewAdded character must is combined with ' 1 ' or ' 2 ' 4. ' 11-19 ', ' 21-26 'Res[i]= Res[i-2] + res[i-1]5. ' 27-99 ' not include *0Res[i]= Res[i-1]----------------------------------------------------1. ' 00 'Res[i]= 0, since ' xx ' is invalid and ' 0 'is invalid.----------------------------------------------------6. *0 (Exclude ' 10 ', ' 20 ') Res[i]= 0think the above analysis is too ugly and complex?When a' 0 ' appears, we need to tackle so many different cases. The related program must is very ugly too!Can We find a to avoid the complex checking?Yes!!!! Why not DoThe dynamic prgramming from right to left? ' 0* ' must be invalid forBoth cases, we can directly ignore it.' 01 '. ' 0 ' is invalid, and ' 01 'is also invalid. for(inti = len-1; I >= 0; i--) {    if(S.charat (i)! = ' 0 ') {        ...} If S.charat (i) is not equal to' 0 ', 1. It must could be separately intepreted ' 1 '. ' 9 'is valid.2. It might be valid forInteprrting together. ' One ' _ ' valid ' is. (note:no ' 0 'start)if(S.charat (i)! = ' 0 ') {Dp[i]= Dp[i+1]; if(I < len-1 && Integer.parseint (s.substring (i, i+2)) <= 26) Dp[i]+ = Dp[i+2];} how easy of ThisSolution, right?Skills:how Initiate the value forFirst intepreted character or ' character '. Can Useint[] DP =New int[Len];DP [0] = 1absolutely no!!!, whatifThe last character was ' 0 '. Like ' 20 'and it would BreakThe elegant code structure. How to use a extra element?int[] DP =New int[Len+1];DP [Len]= 1;1. Last characterifThe last element is ' 0 '. Dp[len-1] still equal to 0ifThe last element was not ' 0 '. Dp[i]= Dp[i+1];2. Last characterifThe second to last element is ' 0 'Dp[len-1] still equal to 0ifThe second to last element was not a ' 0 ' and in the valid range ' 11-26 'Dp[i]+ = Dp[i+2] (1); What a great skill!!!How to give the initial value forBoth CaseThrough a additional element, whileKeep the elegant code structure at the same time.

Solution:

 Public classSolution { Public intnumdecodings (String s) {if(s = =NULL|| S.length () = = 0)            return0; intLen =s.length (); int[] DP =New int[Len+1]; Dp[len]= 1;  for(inti = len-1; I >= 0; i--) {            if(S.charat (i)! = ' 0 ') {Dp[i]= Dp[i+1]; if(I < len-1 && Integer.parseint (s.substring (i, i+2)) <= 26) Dp[i]+ = Dp[i+2]; }        }        returnDp[0]; }}

[leetcode#91] Decode Ways