[leetcode][91] Decode Ways

91. 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 a non-empty string containing only digits, determine the total number of ways to decode it.

Example 1:

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

Example 2:

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

解析

输入一个字符串，把它按照规则解析成字母，返回有多少种解析结果。想到要用动态规划，但是想了一会儿没想出来递推的条件，遂放弃。。

参考答案（别人写的）

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));
            int second = Integer.valueOf(s.substring(i-2, i));
            if(first >= 1 && first <= 9) {
               dp[i] += dp[i-1];  
            }
            if(second >= 10 && second <= 26) {
                dp[i] += dp[i-2];
            }
        }
        return dp[n];
    }
}

关键点在于递推：

int first = Integer.valueOf(s.substring(i-1, i));
int second = Integer.valueOf(s.substring(i-2, i));
if(first >= 1 && first <= 9) {
   dp[i] += dp[i-1];  
}
if(second >= 10 && second <= 26) {
    dp[i] += dp[i-2];
}

仔细推敲这段代码即可。