91. Decode Ways(Leetcode每日一题-2021.04.21)

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Problem

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.

Constraints:

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

Example1

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

Example2

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

Example3

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.

Example4

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

Solution

class Solution {
public:int numDecodings(string s) {int str_len = s.length();vector<int> dp(str_len+1); //dp[i]表示前i个字符解码得到的字符串的个数dp[0] = 1;for(int i = 1;i<=str_len;++i){if(s[i-1] != '0')//如果最后一个字符由s[i]解码dp[i] += dp[i-1];if(i >= 2)//如果最后一个字符由s[i-1]和s[i]解码{int num = (s[i-2] - '0') * 10 + s[i-1]-'0';if(num >= 10 && num <=26)dp[i] += dp[i-2];}}return dp[str_len];}
};