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LintCode Decode Ways

LintCode Decode Ways

作者: Arnold134777 | 来源:发表于2016-03-27 22:14 被阅读75次

    A message containing letters from A-Z is being encoded to numbers using the following mapping:

    'A' -> 1
    'B' -> 2
    ...
    'Z' -> 26
    

    样例

    Given encoded message 12, it could be decoded as AB (1 2) or L (12).
    The number of ways decoding 12 is 2.

    Given an encoded message containing digits, determine the total number of ways to decode it.

    public class Solution {
        /**
         * @param s a string,  encoded message
         * @return an integer, the number of ways decoding
         */
       public int numDecodings(String s) {
            if(null == s || s.length() <= 0)
                return 0;
            
            if(s.length() == 1 && s.charAt(0) == '0')
                return 0;
            
            int[] numDecodings = new int[s.length() + 1];
            numDecodings[0] = 1;
            for(int i = 1;i <= s.length();i++)
            {
                if(s.charAt(i - 1) == '0')
                {
                    if(i- 2 >= 0)
                    {
                        // 如果出现类似"50"这样的数字直接返回0
                        if(s.charAt(i - 2) != '1' && s.charAt(i - 2) != '2')
                        {
                            return 0;
                        }
                        else
                        {
                            numDecodings[i] += numDecodings[i - 2];
                            continue;
                        }
                    }
                }
                numDecodings[i] += numDecodings[i - 1];
                if(i - 2 >= 0)
                {
                    if((s.charAt(i - 2) == '1') || (s.charAt(i - 2) == '2' 
                            && s.charAt(i - 1) >= '1' && s.charAt(i - 1) <= '6'))
                    {
                        numDecodings[i] += numDecodings[i - 2];
                    }
                }
            }
            return numDecodings[s.length()];
        }
    }
    

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