【LeetCode 639】 Decode Ways II

题目描述

A message containing letters from A-Z is being encoded to numbers using the following mapping way:

‘A’ -> 1
‘B’ -> 2
…
‘Z’ -> 26
Beyond that, now the encoded string can also contain the character ‘*’, which can be treated as one of the numbers from 1 to 9.

Given the encoded message containing digits and the character ‘*’, return the total number of ways to decode it.

Also, since the answer may be very large, you should return the output mod 109 + 7.

Example 1:

Input: "*"
Output: 9
Explanation: The encoded message can be decoded to the string: "A", "B", "C", "D", "E", "F", "G", "H", "I".

Example 2:

Input: "1*"
Output: 9 + 9 = 18

Note:

The length of the input string will fit in range [1, 105].
The input string will only contain the character ‘*’ and digits ‘0’ - ‘9’.

思路

动态规划。前一个字符和后一个字符能组成多少种情况，得到当前种类数和之前种类数的关系。有*号的时候关系是乘法，不是加法。
查了好久的if。badcase调试=。=

代码

class Solution {
public:
    int numDecodings(string s) {
        if (s[0] == '0') return 0;
        int n = s.length();
        vector<long long> dp(n+1, 0);
        if (s[0] == '*') dp[0] = 9;
        else if (s[0]-'0' >= 1 && s[0]-'0' <= 9) dp[0] = 1;
        int MOD = 1e9+7;
        
        for (int i=1; i<n; ++i) {
            if (s[i] == '*') {
                if (s[i-1] == '*') {
                    dp[i] = dp[i-1] * 9 + (i > 1 ? dp[i-2] : 1) * 15;
                }else {
                    int num = s[i-1] - '0';
                    if (num == 0) dp[i] = dp[i-1] * 9;
                    else if (num == 1) dp[i] = dp[i-1] * 9 + (i>1 ? dp[i-2] : 1) * 9;
                    else if (num == 2) {
                        dp[i] = dp[i-1] * 9 + (i>1 ? dp[i-2] : 1) * 6;
                    }
                    else dp[i] = dp[i-1] * 9;
                }
            }else {
                if (s[i-1] == '*') {
                    int num = s[i] - '0';
                    if (num == 0) {
                        dp[i] = (i>1 ? dp[i-2] : 1) * 2; 
                    }else {
                        dp[i] = dp[i-1] + (i>1 ? dp[i-2] : 1) * (num <= 6 ? 2 : 1);
                    }
                }else {
                    int num1 = s[i-1] - '0';
                    int num2 = s[i] - '0';
                    if (num1 == 0 && num2 != 0) {
                        dp[i] = dp[i-1];
                    }else {
                        int num = num1 * 10 + num2;
                        if (num >= 10 && num <= 26) {
                            if (num2 != 0) dp[i] = dp[i-1] + (i>1 ? dp[i-2] : 1);
                            else dp[i] = (i > 1 ? dp[i-2] : 1);
                        }else if (num2 != 0) {
                            dp[i] = dp[i-1];
                        }
                    }
                } 
            }
            dp[i] = dp[i] % MOD;
        }
        return dp[n-1];
    }
};