91. Decode Ways（解码方式）

题目链接

https://leetcode.com/problems/decode-ways/description/

题目描述

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.

题目分析

这道题大体思路很简单，主要问题是0的处理：
对于s中的第i个字符s[i]，若s[i]不是0，那么s[i]可以作为一个单独的符号解码；若s[i]是0，那么s[i]必须和s[i-1]组合起来解码。
如果s[i-1]和s[i]组合起来属于[10, 26]，那么可以作为2位数解码， 反之不能。

方法：动态规划

算法描述

遍历s，对s[i]：

s[i] = 0 && 10 <= s[i-1]s[i] <= 26: n[i] = n[i - 2]
1 <= s[i] <= 9 && 10 <= s[i-1]s[i] <= 26: n[i] = n[i - 1] + n[i - 2]
1 <= s[i] <= 9 && !(10 <= s[i-1]s[i] <= 26): n[i] = n[i - 1]
其他情况说明不能解码，直接返回0

参考代码

class Solution {public:    int numDecodings(string s) {        if (s.empty())            return 0;        vector<int> n(s.size() + 1, 1);        if (s[0] != '0')            n[1] = 1;        else            return 0;        for (int i = 2; i < s.size() + 1; i++) {            if (s[i - 1] != '0') {                n[i] = n[i - 1];                if ((s[i - 2] - '0') * 10 + s[i - 1] - '0' <= 26 && s[i - 2] != '0')                    n[i] += + n[i - 2];            } else if (s[i - 2] == '0') {                return 0;            } else {                if ((s[i - 2] - '0') * 10 + s[i - 1] - '0' <= 26)                    n[i] = n[i - 2];                else                    return 0;            }        }        return n[s.size()];    }};