Problem

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.

Note

用parseInt将子字符串转化为int，参见parseInt和valueOf的；

然后用动规方法：dp[i]表示字符串s的前i位（0到i-1）包含decode方法的个数。若前一位i-2到当前位i-1的两位字符串s.substring(i-2, i)对应的数字lastTwo在10到26之间，则当前位dp[i]要加上这两位字符之前一个字符对应的可能性：dp[i-2];若当前位i-1的字符对应1到9之间的数字（不为0），则当前dp[i]要加上前一位也就是第i-2位的可能性：dp[i-1]。最后返回对应字符串s末位的动规结果dp[n]。

Solution

updated 2018-9

class Solution {    public int numDecodings(String s) {        if (s == null || s.length() == 0) return 0;        int n = s.length();        int[] dp = new int[n+1];        dp[0] = 1;                                                  //if the first two digits in [10, 26]        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];             //XXX5X            int lastTwo = Integer.parseInt(s.substring(i-2, i));            if (lastTwo >= 10 && lastTwo <= 26) dp[i] += dp[i-2];   //XXX10X        }        return dp[n];    }}