19.2.23 [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 a non-empty string containing only digits, determine the total number of ways to decode it.

Example 1:

Input: "12"
Output: 2
Explanation: It could be decoded as "AB" (1 2) or "L" (12).

Example 2:

Input: "226"
Output: 3
Explanation: It could be decoded as "BZ" (2 26), "VF" (22 6), or "BBF" (2 2 6).

题意

输出一串数字可能的解读方式，这串数字的解读只能由1~26组成

题解

记忆性递归，好像挺慢的

class Solution {
public:
    vector<int>mark;
    int cnt(string s) {
        int one = s[0] - '0', two = 0, ans = 0, l = s.length();
        if (one == 0)return 0;
        if (mark[l] != -1)return mark[l];
        if (l > 1)two = one * 10 + s[1] - '0';
        else if (l == 0)return 1;
        ans += cnt(s.substr(1));
        if (two <= 26 && two >= 1)
            ans += cnt(s.substr(2));
        mark[l] = ans;
        return ans;
    }
    int numDecodings(string s) {
        int l = s.length();
        mark = vector<int>(l + 1, -1);
        return cnt(s);
    }
};

于是dp，o(n)的

class Solution {
public:
    int numDecodings(string s) {
        int l = s.length();
        vector<int>dp(l + 1, 0);
        dp[0] = 1;
        if (s[0] - '0' != 0)dp[1] = 1;
        for (int i = 1; i < l; i++) {
            int one = s[i - 1] - '0', two = (s[i] - '0') + one * 10;
            if (one != 0 && two <= 26 && two >= 1)
                dp[i+1] +=dp[i - 1];
            if(s[i]!='0')
                dp[i+1] += dp[i];
        }
        return dp[l];
    }
};