leetcode: Decode ways

最新推荐文章于 2019-01-13 06:58:30 发布

原创最新推荐文章于 2019-01-13 06:58:30 发布 · 464 阅读

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本文介绍了一种用于解码数字字符串的算法，该算法能够计算出一个由A到Z字母编码成数字的字符串有多少种不同的解码方式。例如，对于字符串12，它可以被解码为AB或者L，因此总共有2种解码方式。

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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.

class Solution {
public:

    int numDecodingsInt(char * s)
    {
        if (s == NULL)
            return 0;
            
        int strSize = strlen(s);
            
        if (s[0] == '0')
            return 0;
            
        if (strSize == 1)
        {
            return 1;
        }
            
        if (strSize == 2)
        {
            if (s[1] == '0') 
            {
                if (s[0] >= '3')  //30 40 50
                    return 0;
                else
                    return 1;  //10 20
            }
            else if (s[0] == '1')
            {
                return 2;
            }
            else if (s[0] == '2' && s[1] <= '6')
            {
                return 2;
            }
            else
            {
                return 1;
            }
        }      
    }

    int numDecodings(string s) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (s.empty())
            return 0;
             
        int strSize = s.size();
       
        int num1 = numDecodingsInt(&s[strSize-1]); 
        if (strSize == 1)
            return num1;
            
        int num2 = numDecodingsInt(&s[strSize-2]);
        if (strSize == 2)
            return num2;
        
        int curNum = 0;
        
        for (int i = strSize-3; i >= 0; i--)
        {  
            if (s[i] == '0')
            {
                curNum = 0;
            }
            else if (s[i] == '1')
            {
                if (s[i+1] == '0')
                {
                    curNum = num1; //10
                }
                else
                {
                    curNum = num1+num2; //11,12,13,....19
                }
            }
            else if (s[i] == '2')
            {
                if (s[i+1] == '0')
                {
                    curNum = num1; //20
                }
                else if (s[i+1] > '6' && s[i+1] <= '9') //27,28,29
                {
                    curNum = num2;
                }
                else
                {
                    curNum = num1+num2; //21,22,23,24,25,26
                }
            }
            else
            {
                curNum = num2; //31,32....99
            }

            num1 = num2;
            num2 = curNum;               
        }
        
        return curNum;
    }
};