题目：
A message containing letters from A-Z is 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 "12" is 2.
思路：
分情况讨论
1. 当前字符为0
1.1 前一个字符为0,3-9.
1.2 前一个字符为1,2
2. 当前字符非0
2.1 前一个字符为0,3-9
2.2 前一个字符为2，当前字符为7-9
2.3 余下情况
package dp; public class DecodeWays { public int numDecodings(String s) { int n; if (s == null || (n = s.length()) == 0) return 0; int[] dp = new int[n + 1]; dp[0] = 1; if (s.charAt(0) == ‘0‘) return 0; dp[1] = 1; for (int i = 2; i <= n; ++i) { if (s.charAt(i - 1) == ‘0‘) { if (s.charAt(i - 2) != ‘1‘ && s.charAt(i - 2) != ‘2‘) return 0; dp[i] = dp[i - 2]; } else if (s.charAt(i - 2) == ‘0‘ || s.charAt(i - 2) > ‘2‘ || (s.charAt(i - 2) == ‘2‘ && s.charAt(i - 1) > ‘6‘)) { dp[i] = dp[i - 1]; } else { dp[i] = dp[i - 1] + dp[i - 2]; } } return dp[n]; } public static void main(String[] args) { // TODO Auto-generated method stub DecodeWays d = new DecodeWays(); System.out.println(d.numDecodings("10") == 1); System.out.println(d.numDecodings("101") == 1); System.out.println(d.numDecodings("1001") == 0); System.out.println(d.numDecodings("10012") == 0); System.out.println(d.numDecodings("0012") == 0); System.out.println(d.numDecodings("12") == 2); System.out.println(d.numDecodings("128") == 2); System.out.println(d.numDecodings("27") == 1); System.out.println(d.numDecodings("99") == 1); } }
原文：http://www.cnblogs.com/shuaiwhu/p/5098223.html