LeetCode 91. Decode Ways--动态规划DP的Python和Java解法

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本文介绍了一种使用动态规划算法解决数字字符串解码问题的方法。具体地，文章提供了Python和Java两种语言的实现代码，旨在求出给定数字字符串的所有可能解码方式的数量。输入为仅包含数字的非空字符串，输出则是解码的不同方式总数。

题目地址：

LeetCode 动态规划（Dynamic programming）系列题目:LeetCode 动态规划（Dynamic programming）系列题目

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).

python的动态规划解法：

class Solution:
    def numDecodings(self, s):
        dpi_1=0
        dp_i=int(s>'')
        before=''
        for d in s:
             dpi_1, dp_i, before = dp_i, (d>'0')*dp_i + (9<int(before+d)<27)*dpi_1, d
        return dp_i

java的解法：

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];
    }
}