[刷题防痴呆] 0091 - 解码方法 (Decode Ways

题目地址

https://leetcode.com/problems/decode-ways/

题目描述

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 answer is guaranteed to fit 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 = "0"
Output: 0
Explanation: There is no character that is mapped to a number starting with 0.
The only valid mappings with 0 are 'J' -> "10" and 'T' -> "20", neither of which start with 0.
Hence, there are no valid ways to decode this since all digits need to be mapped.
Example 4:

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

思路

• 简单来说, dp[i] = dp[i - 1] + dp[i - 2].
• 但是, 只有当上一位不是0的时候, dp[i]才能继承dp[i - 1].
• 只有当上一位加当前位是在10和26之间的时候, dp[i]才能加上dp[i- 2].

关键点

代码

• 语言支持：Java

class Solution {
    public int numDecodings(String s) {
        if (s.charAt(0) == '0') {
            return 0;
        }
        int n = s.length();
        int[] dp = new int[n + 1];
        dp[0] = 1;
        dp[1] = 1;
        for (int i = 2; i <= n; i++) {
            if (s.charAt(i - 1) != '0') {
                dp[i] = dp[i - 1];
            }
            int preCur = Integer.valueOf(s.substring(i - 2, i));
            if (preCur >= 10 && preCur <= 26) {
                dp[i] += dp[i - 2];
            }
        }
        return dp[n];
    }
}