Decode Ways

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.

 

主要是要区分当最后一位是0时

动态规划，保存没一个字符时的情况

int numDecodings(char* s) {
    int len = strlen(s);
    // if no num or first num is 0, return 0
    if(len == 0 || *s == '0')
        return 0;
    // else open space with len + 1 size ,named num
    int *num = (int *)malloc((len + 1) * sizeof(int));
    num[0] = 1;
    num[1] = 1;
    // go over the string
    for(int i = 1; i < len; i++){
        // if the cur char is 0, 
        if(s[i] == '0'){
            //if the pre char no be 1 or 2 ,return 0
            if(s[i - 1] == '0' || s[i - 1] > '2')
                return 0;
            // else num is pre pre num
            else
                num[i + 1] = num[i - 1];
        }
        // else if the pre char & cur char is between 11 to 26 , the num be the sum of pre and prepre num
        else if(s[i - 1] == '1' || (s[i -1] == '2' && s[i] <= '6'))
            num[i + 1] = num[i] + num[i - 1]; 
        // else other chars while be the previous num
        else
            num[i + 1] = num[i];
    }
    //return last num
    return num[len];
}