Leetcode-Decode Ways

消息解码算法

最新推荐文章于 2020-03-15 15:24:56 发布

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最新推荐文章于 2020-03-15 15:24:56 发布

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原文链接：http://www.cnblogs.com/lishiblog/p/4098786.html

本文介绍了一种使用动态规划解决消息解码问题的方法。对于由数字组成的编码消息，根据A到Z字母与1到26数字的对应关系进行解码。通过动态规划计算不同解码方式的数量，特别注意有效数字范围及特殊情况。

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.

Analysis:

This is a DP problem. d[i] is the number of ways to decode the first i chars.

d[i]=d[i-2]+d[i-1] if num(i-1,i) is a valid number and num(i,i) is a valid number.

NOTE: for a string with 2 chars, the number should be between 10 and 26. For a string with 1 char, the valid number should between 1 and 9. "0","01","02"...like this are all invalid.

Solution:

 1 public class Solution {
 2     public int numDecodings(String s) {
 3         if (s.length()==0) return 0;
 4         int[] d = new int[s.length()+1];
 5         d[0]=1;
 6         for (int i=1;i<d.length;i++){
 7             d[i]=0;
 8             int start = 0;
 9             if (i-2>start) start = i-2;
10             for (int k=start;k<i;k++){
11                String temp = s.substring(k,i);
12                if (temp.length()==1&&!temp.equals("0"))
13                    d[i]+=d[k];
14                if (temp.length()==2){
15                    int num = Integer.parseInt(temp);
16                    if (num<=26 && num>=10)
17                        d[i]+=d[k];
18                }
19             }
20          }
21          return d[s.length()];      
22     }
23 }