[C + +] Leetcode:55 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 " , it could be decoded as " AB "   (1 2) Or " L "   ().

The number of ways decoding is "12" 2.

Ideas:

Set dynamic array dp[i], Dp[i] represents the number of decode means for string s[0,1,2,...., i-1].

Note that if there is a mismatch of 0 in the sequence, then the decoding method is 0, such as the sequence 012,,100 (the second 0 can and 1 composition 10, but the third 0 does not match).

Divided into two cases, consider the end of a number and consider the end of two digits. dynamic Programming equation:

Initial conditions: dp[0] = 1 (according to judging two digits dp[2] + = dp[0] ok), dp[1] = (s[0] = = ' 0 ')? 0:1

Recursive formula: 1. Consider s[0,1,2,..., i-1] At the end of a number as a letter, Dp[i] "1" = s[i-1] = = 0? 0:dp[i-1]

2. Consider s[0,1,2,..., i-1] At the end of two digits as a letter,Dp[i] "2" = IsValid (s[i-2, i-1]) = = true? Dp[i-2]: 0

3. Dp[i] = Dp[i] "1" + Dp[i] "2"

Attention:

1. The setting of the initial conditions dp[0], need to be based on the specific meaning of the code to determine, to meet the actual meaning (dp[2] + = Dp[0], here dp[0] means s[0,1] for a letter, so take 1) .

2. If there is a mismatch of 0 in the sequence, then the decoding method is 0. This is easy to overlook, so be sure to pay attention. If a character is 0, it cannot be decoded, only considering whether it can make a valid character from the previous character.

3. The helper Program (IsValid ()) that adds legal character judgments can make the code more clear. But you can also take a direct judgment. The code is as follows:

for (int i = 2; I <= len; i++)        {            if (s[i-1]! = ' 0 ')                dp[i] = dp[i-1];            else dp[i] = 0;            if (s[i-2] = = ' 1 ' | | (S[i-2] = = ' 2 ' && s[i-1] <= ' 6 '))                Dp[i] + = Dp[i-2];        }

4. The recursive way of dynamic equations, and the meaning of dp[i], need to be accurate and considered comprehensively.

Complexity: O (N)

AC Code:

Class Solution {Public:int numdecodings (string s) {int len = s.length ();                if (len = = 0) return 0;                Dp[i] Indicates the number of decoding methods of s[0,1,..., i-1] int dp[len+1];        Initialize dp[0],dp[1] dp[0] = 1;        if (s[0]! = ' 0 ') {dp[1] = 1;        } else {dp[1] = 0; } for (int i = 2; I <= len; i++) {//If the trailing number is invalid 0, then DP is 0 if (IsValid (S.substr (i-            1, 1))) {Dp[i] = dp[i-1];            } else {Dp[i] = 0;            } if (IsValid (S.substr (i-2, 2))) {Dp[i] + = dp[i-2];    }} return Dp[len]; }public:bool isValid (String str) {//If STR contains invalid 0, such as "02", decoding fails if (str[0] = = ' 0 ') return false        ;        int num = std::stoi (str);       if (num >= 1 && num <=) {return true; } else {return false; }    }};

[C + +] Leetcode:55 Decode Ways