码迷,mamicode.com
首页 > 其他好文 > 详细

Leetcode Decode Ways

时间:2015-11-14 06:25:36      阅读:288      评论:0      收藏:0      [点我收藏+]

标签:

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.


解题思路:

典型的dynamic programming

判断dp[i-1] , dp[i-2] 是否有效

recurrence formula: dp[i] += dp[i-1] if(s(i-1, i) is valid (1-9))  + dp[i-2] (if s(i-2, i)) is valid( 10-26))

base case: dp[0] =1, dp[1] = 1(s(0) is valid) or 0 (s(0) is not valid)


Java code:

public class Solution {
    public int numDecodings(String s) {
        if(s == null || s.length() == 0 || s.equals("0")){
            return 0;
        }
        int[] dp = new int[s.length()+1];
        //base case: dp[0] dp[1]
        dp[0] = 1;
        if(isValid(s.substring(0,1))){
            dp[1] = 1;
        }else {
            dp[1] = 0;
        }
        for(int i = 2; i<= s.length(); i++){
            if(isValid(s.substring(i-1, i))){
                dp[i] += dp[i-1];
            }
            if(isValid(s.substring(i-2,i))){
                dp[i] += dp[i-2];
            }
        }
        return dp[s.length()];
    }
    
    public boolean isValid(String s){
        if(s.charAt(0) == ‘0‘) {
            return false;
        }
        int value = Integer.parseInt(s); // Integer.parseInt: translate string to int
        return value >=1 && value <= 26;
    }
}

Reference:

1. http://www.programcreek.com/2014/06/leetcode-decode-ways-java/

2. http://www.cnblogs.com/springfor/p/3896162.html

 

Leetcode Decode Ways

标签:

原文地址:http://www.cnblogs.com/anne-vista/p/4963741.html

(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!