# 91 Decode Ways

Description

A message containing letters from A-Z can be encoded into numbers using the following mapping:

‘A’ -> “1”
‘B’ -> “2”
…
‘Z’ -> “26”

To decode an encoded message, all the digits must be grouped then mapped back into letters using the reverse of the mapping above (there may be multiple ways). For example, “11106” can be mapped into:

“AAJF” with the grouping (1 1 10 6)
“KJF” with the grouping (11 10 6)
Note that the grouping (1 11 06) is invalid because “06” cannot be mapped into ‘F’ since “6” is different from “06”.

Given a string s containing only digits, return the number of ways to decode it.

The answer is guaranteed to fit in a 32-bit integer.

Examples

Example 1:

Input: s = “12”
Output: 2
Explanation: “12” could be decoded as “AB” (1 2) or “L” (12).

Example 2:

Input: s = “226”
Output: 3
Explanation: “226” could be decoded as “BZ” (2 26), “VF” (22 6), or “BBF” (2 2 6).

Example 3:

Input: s = “0”
Output: 0
Explanation: There is no character that is mapped to a number starting with 0.
The only valid mappings with 0 are ‘J’ -> “10” and ‘T’ -> “20”, neither of which start with 0.
Hence, there are no valid ways to decode this since all digits need to be mapped.

Example 4:

Input: s = “06”
Output: 0
Explanation: “06” cannot be mapped to “F” because of the leading zero (“6” is different from “06”).

Constraints:

1 <= s.length <= 100
s contains only digits and may contain leading zero(s).

思路

是动态规划的思路
假设现在有一串数字

1	2	2	3

初始化1和12所可能做成的count数目

1	2	2	3
1（“1”）	2（“1”“2”/“12”）

对于第三个数字2来说，他能组成的数字是count(“1”) + "22"以及count(“12”) + "2"两种情况。也就是说，假设"112"的count初始为0：

如果"22"能成功被解码，则count+=count(“1”)
如果"2"能成功被解码，则count+=count(“12”)

那么对于该例子来说

1	2	2	3
1（“1”）	2（“1”“2”/“12”）	3 = 1(count(“1”)[“22”=true]) + 2(count(“12”)[“2”=true])

再去看最后一位数

1	2	2	3
1	2	3	5=2(count(“12”)[“23”=true]) + 3(count(“122”)[“3”=true])

最后明确一下初始情况就可以了
几个需要注意的初始样例

“10”
“301”
“27”

代码

class Solution {public int numDecodings(String s) {if(s.length() == 0){return 0;}if(s.charAt(0) == '0'){return 0;}if(s.length() == 1){return 1;}int first = 1;int second = 2;if(s.charAt(1) == '0' || Integer.parseInt(s.substring(0, 2)) > 26){second = 1;}if(s.charAt(1) == '0' && Integer.parseInt(s.substring(0, 2)) > 26){second = 0;}if(s.length() == 2){return second;}for(int i = 2; i < s.length(); i++){int count = 0;if(Integer.parseInt(s.substring(i - 1, i + 1)) < 27 && s.charAt(i - 1) != '0'){count += first;}   if(s.charAt(i) != '0'){count += second;}first = second;second = count;}return second;}
}

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# 91 Decode Ways

Description

Examples

思路

代码

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