【leetcode】91. Decode Ways A-Z的字母表示1-26的数字，反向破解多少种字符串的可能性

1. 题目

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.

2. 思路

递归的方式，直接做会超时，加入了cache后就ok了。
第一个字符作为单一的path数量，与前两个字符作为单一字符的path数量之和，即为总的path数量。

3. 代码

耗时：3ms

class Solution {
public:
    int numDecodings(string s) {
        int len = s.length();
        for (int i = 0; i < len+2; i++) { num_cache[i] = -1; }
        return numDecodings(s, 0);
    }
    
    int numDecodings(string& s, int start) {
        int ls = s.length();
        if (ls == 0) { return 0; }
        int ret = 1;  // 前面是乘法过来的，当为空的时候，这里的返回值应该是1.
        if (start < ls) {
            int num1 = 0;
            int num2 = 0;
            if (numDecodings1(s, start) != 0) {
                if (num_cache[start+1] != -1) {
                    num1 = num_cache[start+1];
                } else {
                    num1 = numDecodings(s, start+1);
                    num_cache[start+1] = num1;
                }
            }
            if (numDecodings2(s, start) != 0) {
                if (num_cache[start+2] != -1) {
                    num2 = num_cache[start+2];
                } else {
                    num2 = numDecodings(s, start+2);
                    num_cache[start+2] = num2;
                }
            }
            ret = num1+num2;
        }
        //cout << "num s=" << s << " start=" << start << " num=" << ret << endl;
        return ret;
    }
    
    int numDecodings1(string& s, int start) {
        int ret = 0;
        if (start < s.length()) {
            ret = s[start] == '0' ? 0 : 1;
        }
        //cout << "num1 s=" << s << " start=" << start << " num=" << ret << endl;
        return ret;
    }
    
    int numDecodings2(string& s, int start) {
        int ret = 0;
        if (start < s.length() - 1) {
            if (s[start] == '1' || s[start] == '2' && s[start+1] < '7') { ret = 1; }
        }
        //cout << "num2 s=" << s << " start=" << start << " num=" << ret << endl;
        return ret;
    }
private:
    int num_cache[10240];
};