91. Decode Ways

A message containing letters from A-Z can be encoded into numbers using the following mapping:

'A' -> "1"
'B' -> "2"
...
'Z' -> "26"

To decode an encoded message, all the digits must be grouped then mapped back into letters using the reverse of the mapping above (there may be multiple ways). For example, "11106" can be mapped into:

• "AAJF" with the grouping (1 1 10 6)
• "KJF" with the grouping (11 10 6)

Note that the grouping (1 11 06) is invalid because "06" cannot be mapped into 'F' since "6" is different from "06".

Given a string s containing only digits, return the number of ways to decode it.

The test cases are generated so that the answer fits in a 32-bit integer.

Example 1:

Input: s = "12"
Output: 2
Explanation: "12" could be decoded as "AB" (1 2) or "L" (12).

Example 2:

Input: s = "226"
Output: 3
Explanation: "226" could be decoded as "BZ" (2 26), "VF" (22 6), or "BBF" (2 2 6).

Example 3:

Input: s = "06"
Output: 0
Explanation: "06" cannot be mapped to "F" because of the leading zero ("6" is different from "06").

Constraints:

• 1 <= s.length <= 100
• s contains only digits and may contain leading zero(s).

Solution:

class Solution {
    public int numDecodings(String s) {
        // DP array to store the subproblem results
        int[] dp = new int[s.length() + 1];  // 226     dp[_, _ , _, _]
        dp[0] = 1; // 表示空字符串的解码方式数量，设为1，因为一个空字符串可以有一种解码方式，即什么也不做
        //  dp[i] 表示字符串 s 的前 i 个字符可以被解码的不同方式的总数。
        // 这里的 i 是从 1 开始计数的，对应到字符串 s 的索引是从 0 到 i-1

        // Ways to decode a string of size 1 is 1. Unless the string is '0'.
        // '0' doesn't have a single digit decode.
        if (s.charAt(0) == '0'){
            dp[1] = 0;
        }else{
            dp[1] = 1;
        }
        // 012
        // 226
        //  | 
        // dp[1, 1, 2,2]
        // 

        for(int i = 2; i < dp.length; i++) {
            // i  = 3 
            // Check if successful single digit decode is possible.
            if (s.charAt(i - 1) != '0') {// 2-1 =1  // 6
               dp[i] = dp[i - 1];   // dp[2] = dp[i-1] = dp[2-1] = dp[1] = 1
              // dp[4] = dp[i-1] =2 
            }

            // Check if successful two digit decode is possible.
            int twoDigit = Integer.valueOf(s.substring(i - 2, i));
            //s.substring() : [i-2,i] = (2 -2 ,i) = 0, 2    22
            // 3 -2 , 3. 1, 3, 1,2    26
            if (twoDigit >= 10 && twoDigit <= 26) { // 
                dp[i] = dp[i] + dp[i - 2]; // dp[2] = dp[2] + dp[i-2] = dp[2-2] = dp[2] + dp[0]
                // 1 + 1 = 2
            }
        }

        return dp[s.length()];
    }
}

// TC: O(n)
// SC: O(n)