[LeetCode]--91. Decode Ways(Python + Java)

See the Problem!

The solution is absolutely Dynamic Programing!
1. Java Solution
Reference:Java clean DP solution with explanation
1.1 use a dp array of size n + 1 to save subproblem solutions. dp[0] means an empty string will have one way to decode, dp[1] means the way to decode a string of size 1. I then check one digit and two digit combination and save the results along the way. In the end, dp[n] will be the end result.

public 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;        dp[1] = s.charAt(0) != '0' ? 1 : 0;        for(int i = 2; i <= n; i++) {            int first = Integer.valueOf(s.substring(i-1, i));            // if first == 0, just dp[i] doesn't change!            if(first >= 1 && first <= 9) {               dp[i] += dp[i-1];              }            if(second >= 10 && second <= 26) {                dp[i] += dp[i-2];            }        }        return dp[n];    }}

1. Python Solution.
   Reference: Python DP solution_by_tusizi
   A digit from index 1 have three condition
   
   • ’?10’ or ‘?20’ this can only divide into ‘10’ or ‘20’ , f(n) = f(n-2)
   • ‘?26’ this can divide into ‘6’ or ‘26’, f(n) = f(n-2)+f(n-1)
   • ‘?09’, ‘?27’ this can only divide into ‘9’ or ‘7’ , f(n) = f(n-1)*
     *

class Solution:# @param s, a string# @return an integerdef numDecodings(self, s):    if not s or s.startswith('0'):        return 0    stack = [1, 1]    for i in range(1, len(s)):        if s[i] == '0':            if s[i-1] == '0' or s[i-1] > '2':  # only '10', '20' is valid                return 0            stack.append(stack[-2])        elif 9 < int(s[i-1:i+1]) < 27:         # '01 - 09' is not allowed            stack.append(stack[-2]+stack[-1])        else:                                  # other case '01, 09, 27'            stack.append(stack[-1])    return stack[-1]