[Leetcode] [Java] Decode Ways

Title:

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.

Test Instructions:

A-Z A message containing letters is encoded into numbers:

' A '-1 ' B '-2 ... ' Z '-26

Given a coded message that contains numbers, it determines how many decoding methods are ultimately in common.

Like what:

give "12", It can be decoded into " AB "   (12) or " L "  

The final decoding method is 2.

Algorithm Analysis:

Reference Blog: http://blog.csdn.net/u011095253/article/details/9248109

* Sweep the string from beginning to end, for example, we want to know, from the first to Dp[i] this bit composed of string, there are a number of decoding combinations, then there are two cases


* First: If dp[i] corresponds to a single character that can be decoded, then Dp[i] includes the number of combinations accumulated by the former dp[i-1] bit dp[i] = dp[i-1]


* Second: If not only dp[i] the corresponding single character can decode, dp[i-1]-dp[i], two characters can also be decoded,

* So not only includes dp[i-1] accumulated number of combinations, also including dp[i-2] bit accumulation of combinations dp[i] = Dp[i-1] + dp[i-2]

* We built an array of n+1, for the sake of brevity, we put a 1 in front. So pay attention to the index in the index-1==string in the array.

AC Code:

<span style= "Font-family:microsoft yahei;font-size:12px;" >public class Solution {public    int numdecodings (String s)    {        int n = s.length ();        if (n==0) return 0;        Int[] dp = new INT[N+1];        Dp[0] = 1;        if (IsValid (s.substring (0,1))) dp[1] = 1;        else dp[1] = 0;        for (int i=2; i<=n;i++)        {            if (IsValid (s.substring (i-1,i)))                dp[i] = dp[i-1];            if (IsValid (s.substring (i-2,i)))                Dp[i] + = Dp[i-2];        }        return dp[n];    }        public Boolean isValid (String s)    {        if (s.charat (0) = = ' 0 ') return false;        int code = Integer.parseint (s);        return code>=1 && code<=26;    }} </span>

Copyright NOTICE: This article is the original article of Bo Master, reprint annotated source

[Leetcode] [Java] Decode Ways