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  • 639. Decode Ways II

    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:

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

    Approach #1: DP. [C++]

    class Solution {
        int mod = 1000000007;
        
        public int numDecodings(String s) {
            if (s.isEmpty()) return 0;
            long[] dp = new long[2];
            dp[0] = 1;
            dp[1] = ways(s.charAt(0));
            // int ans = 0;
            for (int i = 1; i < s.length(); ++i) {
                long ans = ways(s.charAt(i)) * dp[1] + ways(s.charAt(i-1), s.charAt(i)) * dp[0];
                ans %= mod;
                dp[0] = dp[1];
                dp[1] = ans;
            }
            return (int)dp[1];
        }
        
        public int ways(char c) {
            if (c == '*') return 9;
            if (c == '0') return 0;
            return 1;
        }
        
        public int ways(char c1, char c2) {
            if (c1 == '*' && c2 == '*') return 15;
            if (c1 == '*') 
                if (c2 >= '0' && c2 <= '6') return 2;
                else return 1;
            else if (c2 == '*') 
                if (c1 == '1') return 9;
                else if (c1 == '2') return 6;
                else return 0;
            else {
                int num = (c1 - '0') * 10 + (c2 - '0');
                if (num >= 10 && num <= 26) return 1;
                else return 0;
            }
    
        }
    }
    

      

    Reference:

    http://zxi.mytechroad.com/blog/dynamic-programming/leetcode-639-decode-ways-ii/

    永远渴望,大智若愚(stay hungry, stay foolish)
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  • 原文地址:https://www.cnblogs.com/h-hkai/p/10507745.html
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