[leetcode]Decode Ways

Posted 2020-09-20 jzssuanfa

问题描写叙述：

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.

基本思想：

本题利用动态规划记录前面的结果。难点是对0的处理。假设0前面是1或2以外的不论什么数字都无法解码。

代码：

public int numDecodings(String s) {  //java
        //deal with "0"
        if(s.startsWith("0"))
            return 0;
        
        int size = s.length();
        for(int i = 1; i< size; i++)
            if(s.charAt(i) == ‘0‘&&(s.charAt(i-1)>‘2‘||s.charAt(i-1)==‘0‘))
                return 0;
        
        if(size <=1)
            return size;
        
        int top = -1;
        int [] record = new int [size+1];
        record[0] = 0;
        record[1] = 1;
        if(Integer.valueOf(s.substring(0,2)) <=26&&
                Integer.valueOf(s.substring(0,2))>=11&&
                Integer.valueOf(s.substring(0,2))!=20)
            record[2] = 2;
        else record[2] = 1;
        
        for(int i = 3; i <=size; i++){
            if(Integer.valueOf(s.substring(i-2,i))==10 ||
                Integer.valueOf(s.substring(i-2,i))==20)
                record[i] = record[i-2];
            else if(Integer.valueOf(s.substring(i-2,i))<=26&&
                Integer.valueOf(s.substring(i-2,i))>=11)
                record[i] = record[i-2]+record[i-1];
            else record[i] = record[i-1];
        }
        return record[size];
    }