Leetcode - 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.

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.

 

public class Solution {
    // dp[i] = '0' < s.char(i) <= '9' ? d[i - 1] : 0
    //       + 9 < s.sub(i - 2, i) <= 26 ? d[i - 2] : 0
     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; // 注意这个初始化条件
        dp[1] = (s.charAt(0) >= '1' && s.charAt(0) <= '9') ? 1 : 0;
        for (int i = 2; i <= N; i++) {
            char c = s.charAt(i - 1);
            if (c >= '1' && c <= '9')
                dp[i] = dp[i - 1];
            int prev2 = Integer.valueOf(s.substring(i-2, i));
            if (prev2 > 9 && prev2 < 27)
                dp[i] += dp[i - 2];
        }
        return dp[N];
     }
    
    /*Time Limit Exceeded O(N^3)
    public int numDecodings(String s) {
        if (s == null || s.length() == 0)
            return 0;
        int N = s.length();
        int[][] dp = new int[N + 1][N + 1];
        
        // initialization dp[i][i+1]
        for (int i = 0; i < N; i++) {
            char c = s.charAt(i);
            dp[i][i + 1] = (c >= '1' && c <= '9') ? 1 : 0;
        }
        for (int i = 0; i < N - 1; i++) {
            String sub = s.substring(i, i + 2);
            dp[i][i + 2] = (sub.compareTo("10") >= 0 && sub.compareTo("27") < 0) ? 1 : 0;
        }
        for (int k = 3; k <= N; k++) {
            for (int i = 0; i <= N - k ; i++) {
                for (int j = i + 1; j <= i + k - 1; j++) {
                    dp[i][i + k] += dp[i][j] * dp[j][i + k];
                }
            }
        }
        return dp[0][N];
    }
    */
}