[leetcode_91]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.

Initially I used search, which timed out.

Then I realized that if we search from the end, many redundant states can not only be stored but also avoid redundant computation.

Also pay attention to the case where 0 is present.

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class Solution {
public:
    int counts[10000];
    int numDecodings(string s) {
        if(s.length() <= 0 || (s.length() == '1' && s[0] == '0'))return 0;
        if(s[s.length()-1] != '0')counts[s.length() - 1] = 1;
        else counts[s.length()-1] = 0;
        counts[s.length()] = 1;
        for(int i = s.length()-2;i >= 0;i--) {
            if(s[i] == '0') {
                counts[i] = 0;
                continue;
            }
            if(s[i] == '1' || (s[i] == '2'&&s[i+1] <= '6'))counts[i] = counts[i+1] + counts[i+2];
            else
                counts[i] = counts[i+1];
        }
        return counts[0];
    }
};