题目描述：
A message containing letters from A-Z is being encoded to numbers using the following mapping:
'A' -> 1
'B' -> 2
...
'Z' -> 26
Given a non-empty string containing only digits, determine the total number of ways to decode it.
Example 1:
Input: "12"
Output: 2
Explanation: It could be decoded as "AB" (1 2) or "L" (12).
Example 2:
Input: "226"
Output: 3
Explanation: It could be decoded as "BZ" (2 26), "VF" (22 6), or "BBF" (2 2 6).- 正向解法：
public int numDecodings(String s) {int n = s.length();int[] dp = new int[n + 1];dp[0] = 1;dp[1] = s.charAt(0) == '0' ? 0 : 1;for(int i = 2; i <= n; i++){if(s.charAt(i - 1) != '0'){dp[i] += dp[i - 1];}if(s.charAt(i - 2) == '0'){continue;}if(s.charAt(i - 2) == '1' || s.charAt(i - 2) == '2' && s.charAt(i - 1) <= '6'){dp[i] += dp[i - 2];}}return dp[n];}- 逆向解法
public int numDecodings(String s) {int n = s.length();int[] dp = new int[n + 1];dp[n] = 1;dp[n - 1] = s.charAt(n - 1) == '0' ? 0 : 1;for(int i = n - 2; i >= 0; i--){if(s.charAt(i) == '1' || s.charAt(i) == '2' && s.charAt(i + 1) <= '6'){dp[i] += dp[i + 2];}if(s.charAt(i) == '0')continue;elsedp[i] += dp[i + 1];}return dp[0];}