码迷,mamicode.com
首页 > 其他好文 > 详细

[leetcode]Decode Ways

时间:2014-12-15 10:30:25      阅读:196      评论:0      收藏:0      [点我收藏+]

标签:leetcode   算法   

问题描述:

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。如果0前面是1或2以外的任何数字都无法解码。

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];
    }


[leetcode]Decode Ways

标签:leetcode   算法   

原文地址:http://blog.csdn.net/chenlei0630/article/details/41910075

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!