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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.

 

做了好久才AC，主要是脑子不清楚，思路不清晰。虽然最后想通了但花了一整个下午。

 

思路：

f(n) 表示，从第n个字符起的字符串可以排列的方式

从后向前判断，如果该数字是0，则只有0种方式；如果数字非0，则该位数字单独解码，方式有f(n - 1)种。如果该数字可以和它后面的数字一起解码，则该数字和他后面的数字一起，方式再加上f(n-2)种。

class Solution {
public:
    int numDecodings(string s) {
        int iway[2] = {1,1};
        int slen = s.length();
        if(slen <= 0)
            return 0;
        iway[0] = (s[slen - 1] == '0') ? 0 : 1;

        for(int i = s.length() - 2; i >= 0; i--)
        {
            int curWays = 0;
            if(s[i] != '0')
            {
                curWays += iway[0];
            }
            int num = (s[i] - '0') * 10 + (s[i + 1] - '0');
            if(10 <= num && num <= 26)
            {
                curWays += iway[1];
            }
            iway[1] = iway[0];
            iway[0] = curWays;
        }
        return iway[0];
    }
};

大神更加简洁的代码，思路差不多，就是从前向后判断的：

int numDecodings(string s) {
    // empty string or leading zero means no way
    if (!s.size() || s.front() == '0') return 0;

    // r1 and r2 store ways of the last and the last of the last
    int r1 = 1, r2 = 1;

    for (int i = 1; i < s.size(); i++) {
        // zero voids ways of the last because zero cannot be used separately
        if (s[i] == '0') r1 = 0;

        // possible two-digit letter, so new r1 is sum of both while new r2 is the old r1
        if (s[i - 1] == '1' || s[i - 1] == '2' && s[i] <= '6') {
            r1 = r2 + r1;
            r2 = r1 - r2;
        }

        // one-digit letter, no new way added
        else {
            r2 = r1;
        }
    }

    return r1;
}