Decode Ways

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.

Example

Given encoded message 12, it could be decoded as AB (1 2) or L (12).
The number of ways decoding 12 is 2.

第一种笨办法，递归：

public class Solution {
    public int numDecodings(String str) {
        if (str.length() == 0 || str.charAt(0) == '0')  return 0;
        if (str.length() == 1) return 1;

        String subStr = str.substring(0, 2); // 0,1        
        int val = Integer.parseInt(subStr);
        
        if (str.length() == 2) {
            if (val <= 26) {
                return 1 + (str.charAt(1) == '0' ? 0 : 1) ;
            } else {
                return (str.charAt(1) == '0' ? 0 : 1);
            }
        }
        
        if (val <= 26) {
            return numDecodings(str.substring(1)) + numDecodings(str.substring(2));
        } else {
            return numDecodings(str.substring(1));
        }  
    }
}

第二种，DP。

用total[i]表示从i 到 string.length() - 1总共的解法。 我们可以从最后一个数字开始，最后一个数字如果在1-9之间，只有一种decode way, total[length - 1] = 1. 对于倒数第二个数字呢，如果倒数第二个数字不是0，很明显，我们至少有total[i] = total[i + 1], 如果俩个数字合在一起可以组成一个valid number, 那么total[i] += total[i + 2].

public class Solution {
    /**
     * @param s a string,  encoded message
     * @return an integer, the number of ways decoding
     */
    public int numDecodings(String s) {
        if (s == null || s.length() < 1) return 0;
        int[] total = new int[s.length()];

        for (int i = s.length() - 1; i >= 0; i--) {
            char c = s.charAt(i);
            if (c >= '1' && c <= '9') {
                if(i == s.length() - 1) {
                    total[i] = 1;
                } else {
                    total[i] = total[i + 1];
                    String strNumber = s.substring(i, i + 2);
                    int number = Integer.parseInt(strNumber);
                    if (number <= 26) {
                        if (i + 2 == total.length) {
                            total[i] += 1;
                        } else {
                            total[i] += total[i + 2];
                        }
                    }
                }
            }
        }
        return total[0];
    }
}