LeetCode 639. Decode Ways II

原题链接在这里：https://leetcode.com/problems/decode-ways-ii/description/

题目：

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'.

题解：

类似Decode Ways. 也是DP问题. s中含有可表示1~9的*.

若当前是 * , 自己成数有1~9共9种方式. 和前面的char成数看前面char若是'1', 有11~19 共9种方式. 若前面是'2', 有21~26共6种方式. 若前面是*, 共15种方式.

若当前不是*, 自己成数但注意当前是不是'0'. 和前面成数看前面char是不是*. 如果是'*', 当前char又小于等于'6'的话, 需要加上first*2. 以当前char为'5'为例, 有15, 25两种decode方法.

Time Complexity: O(s.length()). Space: O(s.length()).

AC Java:

class Solution {
    int M = 1000000007;
    public int numDecodings(String s) {
        if(s == null || s.length() == 0){
            return 0;
        }
        
        long [] dp = new long[s.length()+1];
        dp[0] = 1;
        dp[1] = s.charAt(0) == '0' ? 0 : s.charAt(0) == '*' ? 9 : 1;
        for(int i = 1; i<s.length(); i++){
            char cur = s.charAt(i);
            char pre = s.charAt(i-1);
            if(cur == '*'){
                // *单独成数
                dp[i+1] = 9*dp[i]%M;
                
                // *和前一位共同成数
                if(pre == '1'){
                    dp[i+1] = (dp[i+1]+9*dp[i-1])%M;
                }else if(pre == '2'){
                    dp[i+1] = (dp[i+1]+6*dp[i-1])%M;
                }else if(pre == '*'){
                    dp[i+1] = (dp[i+1]+15*dp[i-1])%M;
                }
            }else{
                dp[i+1] = cur == '0' ? 0 : dp[i];
                if(pre == '1'){
                    dp[i+1] = (dp[i+1]+dp[i-1])%M;
                }else if(pre == '2' && cur <= '6'){
                    dp[i+1] = (dp[i+1]+dp[i-1])%M;
                }else if(pre == '*'){
                    dp[i+1] = (dp[i+1]+ (cur<='6' ? 2 : 1) *dp[i-1])%M;
                }
            }
        }
        return (int)dp[dp.length-1];
    }
}

更新dp[i+1]时只用到了前两个数. 可以用两个变量代替dp array.

Time Complexity: O(s.length()). Space: O(1).

AC Java:

class Solution {
    int M = 1000000007;
    public int numDecodings(String s) {
        if(s == null || s.length() == 0){
            return 0;
        }
        
        long first = 1;
        long second = s.charAt(0) == '0' ? 0 : s.charAt(0) == '*' ? 9 : 1;
        for(int i = 1; i<s.length(); i++){
            long third = 0;
            char cur = s.charAt(i);
            char pre = s.charAt(i-1);
            if(cur == '*'){
                // *单独成数
                third = 9*second%M;
                
                // *和前一位共同成数
                if(pre == '1'){
                    third = (third+9*first)%M;
                }else if(pre == '2'){
                    third = (third+6*first)%M;
                }else if(pre == '*'){
                    third = (third+15*first)%M;
                }
            }else{
                third = cur == '0' ? 0 : second;
                if(pre == '1'){
                    third = (third+first)%M;
                }else if(pre == '2' && cur <= '6'){
                    third = (third+first)%M;
                }else if(pre == '*'){
                    third = (third+ (cur<='6' ? 2 : 1) *first)%M;
                }
            }
            first = second;
            second = third;
        }
        return (int)second;
    }
}