leetcodeOJ 91. Decode Ways

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参考了大牛的博客：http://blog.csdn.net/linhuanmars/article/details/24570759

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.

主要是分析出递推式：

//思路：动态规划/*递推式： *1.00: ans[i] = 0 *2.11-19、21-26: ans[i] = ans[i-1]+ans[i-2] *3.10、20：ans[i] = ans[i-2] *4.01-09、27-99: ans[i] = ans[i-1] */class Solution {public:    int numDecodings(string s) {        int n = s.size();        if(n == 0 || s[0] == '0')            return 0;                int n1 = 1;         int n2 = 1;        int n3 = 1;        for(int i = 1; i < n; i++){            if(s[i] == '0'){                if(s[i-1] == '1' || s[i-1] == '2')                    n3 = n1;                else                    return 0;            }               else if((s[i-1] >= '3') || (s[i-1] == '0') || (s[i-1] == '2' && s[i] >= '7' && s[i] <= '9')){                n3 = n2;            }            else{                n3 = n1 + n2;            }            n1 = n2;            n2 = n3;        }                return n2;    }};

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